description: Let's define S(x) to be the sum of digits of number x written in decimal system. For example, S(5) = 5, S(10) = 1, S(322) = 7.
  
  We will call an integer x interesting if S(x + 1) < S(x). In each test you will be given one integer n. Your task is to calculate the number of integers x such that 1 ≤ x ≤ n and x is interesting.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 1000) — number of test cases.
  
  Then t lines follow, the i-th line contains one integer n (1 ≤ n ≤ 10^9) for the i-th test case.
  
  Output
  
  Print t integers, the i-th should be the answer for the i-th test case.
  
  Example
  
  Input
  
  
  5
  1
  9
  10
  34
  880055535
  
  
  Output
  
  
  0
  1
  1
  3
  88005553
  
  Note
  
  The first interesting number is equal to 9.
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  public class Main{  
      public static void main (String[] args){
          FastReader s = new FastReader();
          int t=1;t=s.ni();
          for(int test=1;test<=t;test++){int n=s.ni();
          
           System.out.println((n+1)/10);
          }
      }
      
      
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
          public FastReader(){
          br = new BufferedReader(new InputStreamReader(System.in));}
          int ni() { return Integer.parseInt(next()); }
          long nl() { return Long.parseLong(next()); }
          double nd() { return Double.parseDouble(next()); }
          int[] readArray(int n){
  		int[] a = new int[n];
  		for (int i = 0; i < n; i++)
  			a[i] =Integer.parseInt(next());
  		return a;
  	    }
          String next(){
              while (st == null || !st.hasMoreElements()) {try {st = new StringTokenizer(br.readLine());}
              catch (IOException e){e.printStackTrace();}}return st.nextToken();}
              String nextLine(){String str = "";try {str = br.readLine();}catch (IOException e) 
              {e.printStackTrace();}return str;}
      }
  }
  
  
    