#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int k;
    cin >> k;
    long long int i = 0, sum = 0, sumi = 0;
    for (long long int l = 0; l < s.length(); l++) sumi += s[l] - 48;
    ;
    if (sumi <= k) {
      cout << 0 << endl;
      continue;
    }
    while (i < s.length()) {
      sum += (s[i] - 48);
      if (sum >= k) {
        break;
      }
      i++;
    }
    if (i == (s.length() - 1) && sum == k) {
      cout << 0 << endl;
      continue;
    }
    if (i == s.length()) {
      cout << 0 << endl;
      continue;
    }
    string t = "";
    t += '1';
    for (long long int j = i; j < s.length(); j++) {
      t += '0';
    }
    string c = s.substr(i);
    long long int z = stoll(t);
    long long int p = stoll(c);
    long long int ans = z - p;
    cout << ans << endl;
  }
}

### Prompt

Your challenge is to write 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>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int k;
    cin >> k;
    long long int i = 0, sum = 0, sumi = 0;
    for (long long int l = 0; l < s.length(); l++) sumi += s[l] - 48;
    ;
    if (sumi <= k) {
      cout << 0 << endl;
      continue;
    }
    while (i < s.length()) {
      sum += (s[i] - 48);
      if (sum >= k) {
        break;
      }
      i++;
    }
    if (i == (s.length() - 1) && sum == k) {
      cout << 0 << endl;
      continue;
    }
    if (i == s.length()) {
      cout << 0 << endl;
      continue;
    }
    string t = "";
    t += '1';
    for (long long int j = i; j < s.length(); j++) {
      t += '0';
    }
    string c = s.substr(i);
    long long int z = stoll(t);
    long long int p = stoll(c);
    long long int ans = z - p;
    cout << ans << endl;
  }
}

```