A. Div. 7

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

You are given an integer nn. You have to change the minimum number of digits in it in such a way that the resulting number does not have any leading zeroes and is divisible by 77.

If there are multiple ways to do it, print any of them. If the given number is already divisible by 77, leave it unchanged.

Input

The first line contains one integer tt (1≤t≤9901≤t≤990) — the number of test cases.

Then the test cases follow, each test case consists of one line containing one integer nn (10≤n≤99910≤n≤999).

Output

For each test case, print one integer without any leading zeroes — the result of your changes (i. e. the integer that is divisible by 77 and can be obtained by changing the minimum possible number of digits in nn).

If there are multiple ways to apply changes, print any resulting number. If the given number is already divisible by 77, just print it.

Example

input

Copy

3
42
23
377

output

Copy

42
28
777

Note

In the first test case of the example, 4242 is already divisible by 77, so there's no need to change it.

In the second test case of the example, there are multiple answers — 2828, 2121 or 6363.

In the third test case of the example, other possible answers are 357357, 371371 and 378378. Note that you cannot print 077077 or 7777.

Problem solving instructions ： This problem is a math problem , According to the meaning of the topic, change a digit in the number as little as possible , Let the number be 7 to be divisible by , The easiest thing to think of is to change only the lowest position . First, count out the remainder , Compare the remainder with n The lowest point of , Remainder ratio n If the last digit of is small, then directly subtract the remainder , Otherwise, it will be added 7 Re reduction .

#include <stdio.h>

int main(void) 
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int n, k;
		scanf("%d", &n);
		k = n % 7;
		if (n % 10 - k<0)
		{
			n = n + 7 - k;
		}
		else
		{
			n = n - k;
		}
		printf("%d\n", n);
	}
	return 0;
}