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;
int T;
long long n, s;
int a[30], sum[30], nn;
inline long long read() {
  char ch = getchar();
  long long sum = 0;
  bool fu = false;
  while ((ch < '0' || ch > '9') && (ch != '-')) ch = getchar();
  if (ch == '-') fu = true, ch = getchar();
  while (ch >= '0' && ch <= '9') sum = sum * 10 + ch - '0', ch = getchar();
  return (fu ? -sum : sum);
}
int main() {
  T = read();
  while (T--) {
    n = read();
    s = read();
    nn = 0;
    long long tmp = n;
    while (tmp) a[++nn] = tmp % 10, tmp /= 10;
    a[nn + 1] = sum[nn + 1] = 0;
    for (int i = nn; i > 0; i--) sum[i] = sum[i + 1] + a[i];
    if (sum[1] <= s) {
      puts("0");
      continue;
    }
    long long base = 1;
    for (int i = 2; i <= nn + 1; i++) {
      base *= 10;
      if (a[i] == 9) continue;
      if (sum[i] + 1 <= s) {
        printf("%lld\n", (n / base + 1) * base - n);
        break;
      }
    }
  }
  return 0;
}