#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void print(vector<long long> a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << ' ';
  }
  cout << endl;
}
vector<long long> inp(int n) {
  vector<long long> a;
  long long x;
  for (int i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, n, i, j, k, l, c, c1, cnt, flag, m1, m, m2, maxi, mini, x, y, z;
  vector<long long> a, b;
  string s, s1, s2;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> n >> m;
    if (n % m == 0) {
      cout << 0 << endl;
    } else {
      k = m - (n % m);
      cout << k << endl;
    }
  }
}

### 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>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void print(vector<long long> a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << ' ';
  }
  cout << endl;
}
vector<long long> inp(int n) {
  vector<long long> a;
  long long x;
  for (int i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, n, i, j, k, l, c, c1, cnt, flag, m1, m, m2, maxi, mini, x, y, z;
  vector<long long> a, b;
  string s, s1, s2;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> n >> m;
    if (n % m == 0) {
      cout << 0 << endl;
    } else {
      k = m - (n % m);
      cout << k << endl;
    }
  }
}

```