description: This is the easy version of the problem. The difference between the versions is that the easy version does not require you to output the numbers of the rods to be removed. You can make hacks only if all versions of the problem are solved.
  
  Stitch likes experimenting with different machines with his friend Sparky. Today they built another machine.
  
  The main element of this machine are n rods arranged along one straight line and numbered from 1 to n inclusive. Each of these rods must carry an electric charge quantitatively equal to either 1 or -1 (otherwise the machine will not work). Another condition for this machine to work is that the sign-variable sum of the charge on all rods must be zero.
  
  More formally, the rods can be represented as an array of n numbers characterizing the charge: either 1 or -1. Then the condition must hold: a_1 - a_2 + a_3 - a_4 + … = 0, or ∑_{i=1}^n (-1)^{i-1} ⋅ a_i = 0.
  
  Sparky charged all n rods with an electric current, but unfortunately it happened that the rods were not charged correctly (the sign-variable sum of the charge is not zero). The friends decided to leave only some of the rods in the machine. Sparky has q questions. In the ith question Sparky asks: if the machine consisted only of rods with numbers l_i to r_i inclusive, what minimal number of rods could be removed from the machine so that the sign-variable sum of charges on the remaining ones would be zero? Perhaps the friends got something wrong, and the sign-variable sum is already zero. In that case, you don't have to remove the rods at all.
  
  If the number of rods is zero, we will assume that the sign-variable sum of charges is zero, that is, we can always remove all rods.
  
  Help your friends and answer all of Sparky's questions!
  
  Input
  
  Each test contains multiple test cases.
  
  The first line contains one positive integer t (1 ≤ t ≤ 10^3), denoting the number of test cases. Description of the test cases follows.
  
  The first line of each test case contains two positive integers n and q (1 ≤ n, q ≤ 3 ⋅ 10^5) — the number of rods and the number of questions.
  
  The second line of each test case contains a non-empty string s of length n, where the charge of the i-th rod is 1 if s_i is the "+" symbol, or -1 if s_i is the "-" symbol.
  
  Each next line from the next q lines contains two positive integers l_i ans r_i (1 ≤ l_i ≤ r_i ≤ n) — numbers, describing Sparky's questions.
  
  It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5, and the sum of q over all test cases does not exceed 3 ⋅ 10^5.
  
  Output
  
  For each test case, print a single integer — the minimal number of rods that can be removed.
  
  Example
  
  Input
  
  
  3
  14 1
  +--++---++-++-
  1 14
  14 3
  +--++---+++---
  1 14
  6 12
  3 10
  4 10
  +-+-
  1 1
  1 2
  1 3
  1 4
  2 2
  2 3
  2 4
  3 3
  3 4
  4 4
  
  
  Output
  
  
  2
  2
  1
  0
  1
  2
  1
  2
  1
  2
  1
  1
  2
  1
  
  Note
  
  In the first test case for the first query you can remove the rods numbered 5 and 8, then the following set of rods will remain: +--+--++-++-. It is easy to see that here the sign-variable sum is zero.
  
  In the second test case:
  
    * For the first query, we can remove the rods numbered 1 and 11, then the following set of rods will remain: --++---++---. It is easy to see that here the sign-variable sum is zero. 
    * For the second query we can remove the rod numbered 9, then the following set of rods will remain: ---++-. It is easy to see that here the variable sum is zero. 
    * For the third query we can not remove the rods at all. 
solution: import static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  /*
  -> Give your 100%, that's it!
  -> Rules To Solve Any Problem:
      1. Read the problem.
      2. Think About It.
      3. Solve it!
  */
  
  public class Template {
      
  	static int mod = 1000000007;
  
  	public static void main(String[] args){
          FastScanner sc = new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int yo = sc.nextInt();
          while (yo-- > 0) {
              int n = sc.nextInt();
              int q = sc.nextInt();
              int[] prefix = new int[n+1];
              char[] arr = sc.next().toCharArray();
              int[] a = new int[n];
              for(int i = 0; i < n; i++){
                  a[i] = arr[i] == '+' ? 1 : -1;
              }
              for(int i = 1; i < n; i+=2){
                  a[i] *= -1;
              }
              // 1-1+1-1
              for(int i = 1; i <= n; i++){
                  prefix[i] = prefix[i-1] + a[i-1];
              }
              while(q-->0){
                  int l = sc.nextInt();
                  int r = sc.nextInt();
                  int sum = prefix[r] - prefix[l-1];
                  if(sum == 0){
                      out.println(0);
                  }
                  else if(sum%2 == 0){
                      out.println(2);
                  }
                  else {
                      out.println(1);
                  }
              }
          }
          out.close();
  	}
      
     
      
      
      /*
      Source: hu_tao
           Random stuff to try when stuck:
              -if it's 2C then it's dp
              -for combo/probability problems, expand the given form we're interested in
              -make everything the same then build an answer (constructive, make everything 0 then do something)
              -something appears in parts of 2 --> model as graph
              -assume a greedy then try to show why it works
              -find way to simplify into one variable if multiple exist
              -treat it like fmc (note any passing thoughts/algo that could be used so you can revisit them)
              -find lower and upper bounds on answer
              -figure out what ur trying to find and isolate it
              -see what observations you have and come up with more continuations
              -work backwards (in constructive, go from the goal to the start)
              -turn into prefix/suffix sum argument (often works if problem revolves around adjacent array elements)
              -instead of solving for answer, try solving for complement (ex, find n-(min) instead of max)
              -draw something
              -simulate a process
              -dont implement something unless if ur fairly confident its correct
              -after 3 bad submissions move on to next problem if applicable
              -do something instead of nothing and stay organized
              -write stuff down
           Random stuff to check when wa:
              -if code is way too long/cancer then reassess
              -switched N/M
              -int overflow
              -switched variables
              -wrong MOD
              -hardcoded edge case incorrectly
           Random stuff to check when tle:
              -continue instead of break
              -condition in for/while loop bad
           Random stuff to check when rte:
              -switched N/M
              -long to int/int overflow
              -division by 0
              -edge case for empty list/data structure/N=1
        */
  
  	public static class Pair {
  		int x;
  		int y;
  
  		public Pair(int x, int y) {
  			this.x = x;
  			this.y = y;
  		}
  	}
  
  	public static void sort(int[] arr) {
  		ArrayList<Integer> ls = new ArrayList<Integer>();
  		for (int x : arr)
  			ls.add(x);
  		Collections.sort(ls);
  		for (int i = 0; i < arr.length; i++)
  			arr[i] = ls.get(i);
  	}
  
  	public static long gcd(long a, long b) {
  		if (b == 0)
  			return a;
  		return gcd(b, a % b);
  	}
  
  	static boolean[] sieve(int N) {
  		boolean[] sieve = new boolean[N + 1];
  		for (int i = 2; i <= N; i++)
  			sieve[i] = true;
  
  		for (int i = 2; i <= N; i++) {
  			if (sieve[i]) {
  				for (int j = 2 * i; j <= N; j += i) {
  					sieve[j] = false;
  				}
  			}
  		}
  		return sieve;
  	}
  
  	public static long power(long x, long y, long p) {
  		long res = 1L;
  		x = x % p;
  		while (y > 0) {
  			if ((y & 1) == 1)
  				res = (res * x) % p;
  			y >>= 1;
  			x = (x * x) % p;
  		}
  		return res;
  	}
  
  	public static void print(int[] arr) {
  		//for debugging only
  		for (int x : arr)
  			out.print(x + " ");
  		out.println();
  	}
  
  	static class FastScanner {
  		private int BS = 1 << 16;
  		private char NC = (char) 0;
  		private byte[] buf = new byte[BS];
  		private int bId = 0, size = 0;
  		private char c = NC;
  		private double cnt = 1;
  		private BufferedInputStream in;
  
  		public FastScanner() {
  			in = new BufferedInputStream(System.in, BS);
  		}
  
  		public FastScanner(String s) {
  			try {
  				in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
  			} catch (Exception e) {
  				in = new BufferedInputStream(System.in, BS);
  			}
  		}
  
  		private char getChar() {
  			while (bId == size) {
  				try {
  					size = in.read(buf);
  				} catch (Exception e) {
  					return NC;
  				}
  				if (size == -1)
  					return NC;
  				bId = 0;
  			}
  			return (char) buf[bId++];
  		}
  
  		public int nextInt() {
  			return (int) nextLong();
  		}
  
  		public int[] readInts(int N) {
  			int[] res = new int[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = (int) nextLong();
  			}
  			return res;
  		}
  
  		public long[] readLongs(int N) {
  			long[] res = new long[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = nextLong();
  			}
  			return res;
  		}
  
  		public long nextLong() {
  			cnt = 1;
  			boolean neg = false;
  			if (c == NC)
  				c = getChar();
  			for (; (c < '0' || c > '9'); c = getChar()) {
  				if (c == '-')
  					neg = true;
  			}
  			long res = 0;
  			for (; c >= '0' && c <= '9'; c = getChar()) {
  				res = (res << 3) + (res << 1) + c - '0';
  				cnt *= 10;
  			}
  			return neg ? -res : res;
  		}
  
  		public double nextDouble() {
  			double cur = nextLong();
  			return c != '.' ? cur : cur + nextLong() / cnt;
  		}
  
  		public double[] readDoubles(int N) {
  			double[] res = new double[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = nextDouble();
  			}
  			return res;
  		}
  
  		public String next() {
  			StringBuilder res = new StringBuilder();
  			while (c <= 32)
  				c = getChar();
  			while (c > 32) {
  				res.append(c);
  				c = getChar();
  			}
  			return res.toString();
  		}
  
  		public String nextLine() {
  			StringBuilder res = new StringBuilder();
  			while (c <= 32)
  				c = getChar();
  			while (c != '\n') {
  				res.append(c);
  				c = getChar();
  			}
  			return res.toString();
  		}
  
  		public boolean hasNext() {
  			if (c > 32)
  				return true;
  			while (true) {
  				c = getChar();
  				if (c == NC)
  					return false;
  				else if (c > 32)
  					return true;
  			}
  		}
  	}
  
  	//	For Input.txt and Output.txt	
  	//	FileInputStream in = new FileInputStream("input.txt");
  	//	FileOutputStream out = new FileOutputStream("output.txt");
  	//	PrintWriter pw = new PrintWriter(out);
  	//	Scanner sc = new Scanner(in);
  }