description: Consider the insertion sort algorithm used to sort an integer sequence [a_1, a_2, …, a_n] of length n in non-decreasing order.
  
  For each i in order from 2 to n, do the following. If a_i ≥ a_{i-1}, do nothing and move on to the next value of i. Otherwise, find the smallest j such that a_i < a_j, shift the elements on positions from j to i-1 by one position to the right, and write down the initial value of a_i to position j. In this case we'll say that we performed an insertion of an element from position i to position j.
  
  It can be noticed that after processing any i, the prefix of the sequence [a_1, a_2, …, a_i] is sorted in non-decreasing order, therefore, the algorithm indeed sorts any sequence.
  
  For example, sorting [4, 5, 3, 1, 3] proceeds as follows: 
  
    * i = 2: a_2 ≥ a_1, do nothing; 
    * i = 3: j = 1, insert from position 3 to position 1: [3, 4, 5, 1, 3]; 
    * i = 4: j = 1, insert from position 4 to position 1: [1, 3, 4, 5, 3]; 
    * i = 5: j = 3, insert from position 5 to position 3: [1, 3, 3, 4, 5]. 
  
  
  
  You are given an integer n and a list of m integer pairs (x_i, y_i). We are interested in sequences such that if you sort them using the above algorithm, exactly m insertions will be performed: first from position x_1 to position y_1, then from position x_2 to position y_2, ..., finally, from position x_m to position y_m.
  
  How many sequences of length n consisting of (not necessarily distinct) integers between 1 and n, inclusive, satisfy the above condition? Print this number modulo 998 244 353.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
  
  The first line of each test case contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5; 0 ≤ m < n) — the length of the sequence and the number of insertions.
  
  The i-th of the following m lines contains two integers x_i and y_i (2 ≤ x_1 < x_2 < … < x_m ≤ n; 1 ≤ y_i < x_i). These lines describe the sequence of insertions in chronological order.
  
  It is guaranteed that the sum of m over all test cases does not exceed 2 ⋅ 10^5. Note that there is no constraint on the sum of n of the same kind.
  
  Output
  
  For each test case, print the number of sequences of length n consisting of integers from 1 to n such that sorting them with the described algorithm produces the given sequence of insertions, modulo 998 244 353.
  
  Example
  
  Input
  
  
  3
  3 0
  3 2
  2 1
  3 1
  5 3
  3 1
  4 1
  5 3
  
  
  Output
  
  
  10
  1
  21
  
  Note
  
  In the first test case, the algorithm performs no insertions — therefore, the initial sequence is already sorted in non-decreasing order. There are 10 such sequences: [1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 2], [1, 2, 3], [1, 3, 3], [2, 2, 2], [2, 2, 3], [2, 3, 3], [3, 3, 3].
  
  In the second test case, the only sequence satisfying the conditions is [3, 2, 1].
  
  In the third test case, [4, 5, 3, 1, 3] is one of the sought sequences.
solution: import java.util.*;
  import java.io.*;
  
  public class C1558 {
  	static int mod = 998244353;
  	static long[] fac, facInv;
  
  	static long modPow(long a, long e) {
  		if (e == 0) {
  			return 1;
  		}
  		long prev = modPow(a, e / 2);
  		if (e % 2 == 0) {
  			return prev * prev % mod;
  		}
  		return prev * prev % mod * a % mod;
  	}
  
  	static long modInv(long x) {
  		return modPow(x, mod - 2);
  	}
  
  	static void preprocess(int n) {
  		fac = new long[n + 1];
  		facInv = new long[n + 1];
  		fac[0] = facInv[0] = 1;
  		for (int i = 1; i <= n; i++) {
  			fac[i] = fac[i - 1] * i % mod;
  			facInv[i] = facInv[i - 1] * modInv(i) % mod;
  		}
  	}
  
  	public static long nck(int n, int k) {
  		if (k > n || k < 0 || n < 0)
  			return 0;
  		return fac[n] * facInv[k] % mod * facInv[n - k] % mod;
  	}
  
  	public static void main(String[] args) throws IOException {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		preprocess((int) 4e5 + 5);
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int m = sc.nextInt();
  			Treap treap = new Treap(1);
  			int pos = 1;
  			int comp = 1;
  			while (m-- > 0) {
  				int x = sc.nextInt();
  				update(treap, x - pos - 1);
  				int y = sc.nextInt();
  
  				if (y == 1) {
  					treap = merge(new Treap(1), treap);
  					comp++;
  //					System.out.println("A" + " " + treap.subtreeSize);
  				} else {
  					Treap[] s0 = split(treap, y - 1);
  					int sizeLeft = sum(s0[0]);
  
  					if (sizeLeft == y - 1) {
  //						System.out.println("B");
  						update(s0[0], 1);
  						treap = merge(s0[0], s0[1]);
  					} else {
  						update(s0[0], y - sizeLeft);
  
  						Treap newNode = new Treap(sizeLeft - y + 1);
  						treap = merge(s0[0], merge(newNode, s0[1]));
  						comp++;
  //						System.out.println("C");
  					}
  				}
  				pos = x;
  			}
  //			pw.println(comp);
  
  			pw.println(nck(2 * n - comp, n - comp));
  		}
  		pw.close();
  	}
  
  	// lower priority on top, all methods return the new treap root
  	// To add new seg-tree supported properties, edit recalc()
  	// To add lazyprop values, edit recalc() and prop()
  
  	// If you only add by merging, skip add() and recalc()
  	// If you don't need lazyprop, skip prop() and rangeAdd()
  	static class Treap {
  		double priority;
  		Treap[] kids = new Treap[2];
  
  		int subtreeSize, data, sum;
  
  		public Treap(int data) {
  			this.data = data;
  			priority = Math.random();
  			recalc(this);
  		}
  
  		// returns lefthalf, rightHalf
  		// nInLeft is size of left treap, aka index of first thing in right trea
  	}
  
  	static Treap[] split(Treap me, int nInLeft) {
  		if (me == null)
  			return new Treap[] { null, null };
  		if (sum(me.kids[0]) >= nInLeft) {
  			Treap[] leftRes = split(me.kids[0], nInLeft);
  			me.kids[0] = leftRes[1];
  			recalc(me);
  			return new Treap[] { leftRes[0], me };
  		} else {
  			nInLeft = nInLeft - sum(me.kids[0]) - me.data;
  			Treap[] rightRes = split(me.kids[1], nInLeft);
  			me.kids[1] = rightRes[0];
  			recalc(me);
  			return new Treap[] { me, rightRes[1] };
  		}
  	}
  
  	static Treap merge(Treap l, Treap r) {
  		if (l == null)
  			return r;
  		if (r == null)
  			return l;
  
  		if (l.priority < r.priority) {
  			l.kids[1] = merge(l.kids[1], r);
  			recalc(l);
  			return l;
  		} else {
  			r.kids[0] = merge(l, r.kids[0]);
  			recalc(r);
  			return r;
  		}
  	}
  
  	static void recalc(Treap me) {
  		if (me == null)
  			return;
  		me.subtreeSize = 1;
  		me.sum = me.data;
  		for (Treap t : me.kids)
  			if (t != null)
  				me.subtreeSize += t.subtreeSize;
  		for (Treap t : me.kids)
  			if (t != null)
  				me.sum += t.sum;
  	}
  
  	static void update(Treap t, int val) {
  		if (t.kids[1] != null) {
  			update(t.kids[1], val);
  		} else {
  			t.data += val;
  		}
  		recalc(t);
  	}
  
  	static int size(Treap t) {
  		return t == null ? 0 : t.subtreeSize;
  	}
  
  	static int sum(Treap t) {
  		return t == null ? 0 : t.sum;
  	}
  
  	static class Scanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public Scanner(InputStream s) {
  			br = new BufferedReader(new InputStreamReader(s));
  		}
  
  		public Scanner(FileReader f) {
  			br = new BufferedReader(f);
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public int[] nextIntArr(int n) throws IOException {
  			int[] arr = new int[n];
  			for (int i = 0; i < n; i++) {
  				arr[i] = Integer.parseInt(next());
  			}
  			return arr;
  		}
  
  	}
  
  }
  