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;
  
  public class Sumanumeros {
  	
  	
  	
  	public static void main(String[] args) {
  	     
  		try {
  				BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  
  				String cant = br.readLine();
  				int n = Integer.parseInt(cant);
  				
  				while(n-- > 0) {
  					int num = Integer.parseInt(br.readLine());
  					int res = num / 10;
  					
  					char [] digitos = String.valueOf(num).toCharArray();
  					if(digitos[digitos.length-1]=='9') {
  						res++;
  					}
  					
  					System.out.println(res);
  				}
  				
  				
              	
  		} catch (IOException e) {
  			e.printStackTrace();
  		}
   }
  /*
  	private static int interesantes(int num) {
  		
  		int cuenta = 0;
  		for(int i=9; i<=num; i++) {
  			if(esInteresante(i)) {
  				cuenta++;
  			}
  		}
  		return cuenta;
  		
  	}
  
  	private static boolean esInteresante(int i) {
  		
  		if(evaluar(i+1) < evaluar(i)) {
  			return true;
  		}else {
  			return false;
  		}
  		
  	}
  
  	private static int evaluar(int n) {
  		
  		char [] digitos = String.valueOf(n).toCharArray();
  		int suma = 0;
  		for(int i=0; i<digitos.length; i++) {
  			suma += Character.getNumericValue(digitos[i]);
  			
  		}
  		
  		return suma;
  	}
  */
  }
  