Problem: You are given a positive integer n. In one move, you can increase n by one (i.e. make n := n + 1). Your task is to find the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains two integers n and s (1 ≤ n ≤ 10^{18}; 1 ≤ s ≤ 162).

Output

For each test case, print the answer: the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

Example

Input


5
2 1
1 1
500 4
217871987498122 10
100000000000000001 1


Output


8
0
500
2128012501878
899999999999999999

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int f();
int main() {
  int q = 0, e = 0;
  int arr[20000], arr1[20000], arr2[20000];
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> arr[i] >> arr1[i];
  }
  for (int i = 0; i < q; i++) {
    if (arr[i] % arr1[i] != 0)
      arr2[i] = arr1[i] - (arr[i] % arr1[i]);
    else
      arr2[i] = 0;
  }
  for (int i = 0; i < q; i++) {
    cout << arr2[i] << "\t";
  }
}
int f() { return 0; }