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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxx = (int)1e7;
long long SOD(long long n) {
  long long ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    vector<int> num;
    long long n, s, nnn;
    cin >> n >> s;
    nnn = n;
    if (SOD(n) <= s) {
      cout << 0 << endl;
      continue;
    }
    while (n) {
      num.push_back(n % 10);
      n /= 10;
    }
    reverse(num.begin(), num.end());
    int sum = 0;
    int start = 0;
    for (int i = 0; i < num.size(); i++) {
      sum += num[i];
      if (sum >= s) {
        if (i != 0 && num[i - 1] != 9) {
          num[i - 1]++;
          start = i;
          break;
        }
        if (i == 0) {
          num.insert(num.begin(), 1);
          start = 1;
          break;
        } else if (num[i - 1] == 9) {
          while (i != 0 && num[i - 1] == 9) {
            i--;
          }
          i--;
        }
      }
    }
    for (int i = start; i < num.size(); i++) {
      num[i] = 0;
    }
    int p = num.size() - 1;
    long long nn = 0;
    for (int i = 0; i < num.size(); i++) {
      nn += num[i] * round(pow(10, p));
      p--;
    }
    cout << (nn - nnn) << endl;
  }
  return 0;
}