#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long t, n;
vector<int> a, b;
long long f(long long s) {
  long long ans = 0;
  while (s) {
    ans += s % 10;
    s /= 10;
  }
  return ans;
}
int main() {
  cin >> t;
  while (t--) {
    int s;
    cin >> n >> s;
    long long q = n;
    long long POW = 1, ans = 0;
    while (f(q) > s) {
      if (q % 10) ans += POW * (10 - q % (10));
      if (q % 10) q += 10;
      q /= 10;
      POW *= 10;
    }
    cout << ans << endl;
  }
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long t, n;
vector<int> a, b;
long long f(long long s) {
  long long ans = 0;
  while (s) {
    ans += s % 10;
    s /= 10;
  }
  return ans;
}
int main() {
  cin >> t;
  while (t--) {
    int s;
    cin >> n >> s;
    long long q = n;
    long long POW = 1, ans = 0;
    while (f(q) > s) {
      if (q % 10) ans += POW * (10 - q % (10));
      if (q % 10) q += 10;
      q /= 10;
      POW *= 10;
    }
    cout << ans << endl;
  }
}

```