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.util.Scanner;

public class Codeforces {
	static Scanner sc=new Scanner(System.in);
	
	public static void main(String[] args) {
		int t=1;
		t=sc.nextInt();
		
		while(t-->0) {
			solve();
		}
		
	}

	private static void solve() {
		
		int n=sc.nextInt();
		if(n<9) {
			System.out.println(0);
		}else if(n==9) {
			System.out.println(1);
		}else {
			System.out.println(n/10);
		}
		
		/*int n=sc.nextInt();
		int arr[]=new int[n];
		for(int i=0;i<n;i++) {
			arr[i]=sc.nextInt();
		}
		
		int sl=Integer.MIN_VALUE;
		int fl=Integer.MIN_VALUE;
		
		for(int i=0;i<n;i++) {
			if(arr[i]>fl) {
				sl=fl;
				fl=arr[i];
			}else if(arr[i]>sl && arr[i]!=fl) {
				sl=arr[i];
			}
		}
		System.out.println(fl+"  "+sl);
		if(sl==Integer.MIN_VALUE) {
			System.out.println(fl*fl);
		}else {
			System.out.println((long)fl*sl);
		}
		
		*/
	}

	/*private static void solve() {
		int l=sc.nextInt();
		int t=sc.nextInt();
		String str=sc.next();//.substring(0,l);
		while(t-->0) {
			int low=sc.nextInt();
			int upper=sc.nextInt();
			if(low<0 || upper>l) {
				return ;
			}
			String subStr=str.substring(low-1,upper);
			
			int count=0;
			for(int i=0;i<subStr.length();i++) {
				count+= subStr.charAt(i)-'a'+1;
			}
			System.out.println(count);
			
			
		}
	}*/
}
