#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
constexpr int MOD = 1e9 + 7;
void solve() {
  string N;
  int S;
  cin >> N >> S;
  int s = 0;
  int idx = -1;
  string orig = N;
  auto sum = [&](string T) {
    int s = 0;
    for (int i = 0; i < T.size(); i++) {
      s += (T[i] - '0');
    }
    return s;
  };
  auto check = [&](string T) {
    int idx = -1;
    int s = 0;
    for (int i = 0; i < T.size(); i++) {
      if (s + (T[i] - '0') > S) {
        idx = i;
        break;
      }
      s += (T[i] - '0');
    }
    if (idx == -1) {
      return T;
    }
    if (idx == 0) {
      string ans;
      ans += '1';
      for (int i = idx; i < T.size(); i++) ans += '0';
      return ans;
    } else {
      T[idx] = '0';
      for (int i = idx + 1; i < T.size(); i++) T[i] = '0';
      idx--;
      while (idx >= 0 && T[idx] == '9') {
        T[idx] = '0';
        idx--;
      }
      if (idx == -1) {
        string ans = '1' + T;
        return ans;
      } else {
        T[idx] = T[idx] + 1;
      }
      return T;
    }
  };
  while (sum(N) > S) {
    N = check(N);
  }
  long long val = stoll(N);
  cout << val - stoll(orig) << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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")
#pragma GCC optimization("unroll-loops")
using namespace std;
constexpr int MOD = 1e9 + 7;
void solve() {
  string N;
  int S;
  cin >> N >> S;
  int s = 0;
  int idx = -1;
  string orig = N;
  auto sum = [&](string T) {
    int s = 0;
    for (int i = 0; i < T.size(); i++) {
      s += (T[i] - '0');
    }
    return s;
  };
  auto check = [&](string T) {
    int idx = -1;
    int s = 0;
    for (int i = 0; i < T.size(); i++) {
      if (s + (T[i] - '0') > S) {
        idx = i;
        break;
      }
      s += (T[i] - '0');
    }
    if (idx == -1) {
      return T;
    }
    if (idx == 0) {
      string ans;
      ans += '1';
      for (int i = idx; i < T.size(); i++) ans += '0';
      return ans;
    } else {
      T[idx] = '0';
      for (int i = idx + 1; i < T.size(); i++) T[i] = '0';
      idx--;
      while (idx >= 0 && T[idx] == '9') {
        T[idx] = '0';
        idx--;
      }
      if (idx == -1) {
        string ans = '1' + T;
        return ans;
      } else {
        T[idx] = T[idx] + 1;
      }
      return T;
    }
  };
  while (sum(N) > S) {
    N = check(N);
  }
  long long val = stoll(N);
  cout << val - stoll(orig) << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

```