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.*;
  import java.lang.*;
  import java.util.*;
  
  public class interesting{
      // static int getSum(int n)
      // {    
      //     int sum = 0;
            
      //     while (n != 0)
      //     {
      //         sum = sum + n % 10;
      //         n = n/10;
      //     }
        
      // return sum;
      // }
      public static void main (String[] args) throws java.lang.Exception
      {
          
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
          int tc = Integer.parseInt(br.readLine());
          while(tc-->0){
              int inter =0;
              StringTokenizer st1 = new StringTokenizer(br.readLine());
              int x1 =Integer.parseInt(st1.nextToken());
              // for(int i=1;i<=x1;i++){
              //     if(getSum(i+1)<getSum(i)){
              //         inter++;
              //     }
              // }
              // writer.write(inter+"");
              if((x1+1)/10<1){
                  writer.write(inter+"\n");
              }
              else writer.write((x1+1)/10+"\n");
          }
          writer.flush();
          writer.close();
          br.close();
      }
  }