#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
const int MOD = 1e9 + 7;
const long double PI = 3.1415926535898;
const long long INF = LLONG_MAX;
int sum(long long a) {
  int res = 0;
  while (a) {
    res += a % 10;
    a /= 10;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long ans = 0;
    long long mul = 1;
    while (sum(n) > s) {
      long long last = n % 10;
      ans += (10LL - last) * mul;
      mul *= 10;
      n /= 10;
      n += 1;
    }
    cout << ans << "\n";
  }
}

### Prompt

In cpp, your task is to solve 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>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
const int MOD = 1e9 + 7;
const long double PI = 3.1415926535898;
const long long INF = LLONG_MAX;
int sum(long long a) {
  int res = 0;
  while (a) {
    res += a % 10;
    a /= 10;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long ans = 0;
    long long mul = 1;
    while (sum(n) > s) {
      long long last = n % 10;
      ans += (10LL - last) * mul;
      mul *= 10;
      n /= 10;
      n += 1;
    }
    cout << ans << "\n";
  }
}

```