#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int MAX = 1000005;
const double PI = acos(-1.0);
int SetBit(int n, int X) { return n | (1 << X); }
int ClearBit(int n, int X) { return n & ~(1 << X); }
int ToggleBit(int n, int X) { return n ^ (1 << X); }
bool CheckBit(int n, int X) { return (bool)(n & (1 << X)); }
int main(void) {
  int tc, a, b, ans;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%d %d", &a, &b);
    ans = a % b == 0 ? 0 : b - (a % b);
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int MAX = 1000005;
const double PI = acos(-1.0);
int SetBit(int n, int X) { return n | (1 << X); }
int ClearBit(int n, int X) { return n & ~(1 << X); }
int ToggleBit(int n, int X) { return n ^ (1 << X); }
bool CheckBit(int n, int X) { return (bool)(n & (1 << X)); }
int main(void) {
  int tc, a, b, ans;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%d %d", &a, &b);
    ans = a % b == 0 ? 0 : b - (a % b);
    printf("%d\n", ans);
  }
  return 0;
}

```