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.
import java.io.*;
import java.util.*;


public class c1553 {

    public static void main(String[] args) throws Exception{
        FastScanner fs = new FastScanner();
        int T = 1;
        T=fs.nextInt();
        for (int tt = 0; tt < T; tt++) {
        		long n = fs.nextLong();
        		long x = n/10;
        		long  m = n%10==9?1:0;
        		System.out.println(x+m);
        }
    }

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a) l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    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());
        }
 
        long[] readArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }
 
        long nextLong() {
            return Long.parseLong(next());
        }
    }

}
