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.util.*;
  import java.io.*;
  public class codeforces {
      public static Boolean helper(char s[],char rs[],int i,int j,int dir){
              if(j==rs.length)
                  return true;
              boolean ok1=false,ok2=false;
              if((i+1)<s.length && s[i+1]==rs[j] && dir==1)
               ok1  =   helper(s,rs,i+1,j+1,1);
              if((i-1)>=0 && s[i-1]==rs[j]){
                  ok2 = helper(s,rs,i-1,j+1,0);
              }
              if(ok1||ok2)
                return true;
              else
                  return false;
      }
      public static void main(String []args){
          FastScanner sc = new FastScanner();
          int t = sc.nextInt();
          while(t-->0)   {
               int n = sc.nextInt();
               int count = n/10;
               if(n%10==9)
                   count++;
               System.out.println(count);
          }
      }
      public static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }
  