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 long double PI = acos(-1);
const long long MOD = 1e9 + 7;
const long long INF = LLONG_MAX;
const long long SIZE = 100000;
void solve() {
  long long n, s;
  cin >> n >> s;
  long long sum = 0, dig = 0, fir;
  long long k = n;
  vector<long long> v;
  while (k != 0) {
    long long rem = k % 10;
    sum += rem;
    fir = rem;
    v.push_back(rem);
    k /= 10;
    dig++;
  }
  if (sum <= s) {
    cout << 0 << endl;
  } else {
    reverse(v.begin(), v.end());
    long long sum2 = 0, fin;
    long long pos = -1;
    for (long long i = 0; i < (long long)v.size(); i++) {
      sum2 += v[i];
      if (sum2 > s) {
        pos = i;
        break;
      }
    }
    sum2 -= v[pos];
    long long yup = -1;
    for (long long i = pos - 1; i >= 0; i--) {
      v[i] += 1;
      sum2 += 1;
      if (sum2 <= s) {
        yup = i;
        break;
      }
      sum2 -= v[i];
    }
    if (yup == -1) {
      long long fin = pow(10, dig);
      cout << fin - n << endl;
    } else {
      for (long long i = yup + 1; i < (long long)v.size(); i++) {
        v[i] = 0;
      }
      long long fin = 0;
      long long len = v.size() - 1;
      for (long long i = 0; i < (long long)v.size(); i++) {
        fin += v[i] * pow(10, len);
        len--;
      }
      cout << fin - n << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}