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 N = 1e5 + 10;
int tl;
int a[500];
long long int judge(long long int t) {
  long long int sum = 0;
  while (t) {
    sum += t % 10;
    t /= 10;
  }
  return sum;
}
long long int power(int t) {
  long long int tot = 1;
  for (int i = 1; i <= t; i++) tot *= 10;
  return tot;
}
void fj(long long int t) {
  tl = 0;
  while (t) {
    a[tl++] = t % 10;
    t /= 10;
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long int n, nl, s, sum = 0;
    scanf("%lld%lld", &n, &s);
    nl = n;
    tl = 0;
    while (nl > 0) {
      a[tl++] = nl % 10;
      nl /= 10;
    }
    int len = tl;
    for (int i = 0; i < len; i++) {
      if (judge(n) <= s) break;
      nl = n;
      fj(nl);
      if (a[i] != 0) {
        n += (10 - a[i]) * power(i);
        sum += (10 - a[i]) * power(i);
      }
    }
    printf("%lld\n", sum);
  }
  return 0;
}