#include <bits/stdc++.h>
int sum(uint64_t v) {
  int s = 0;
  while (v) {
    s += v % 10;
    v -= v % 10;
    v /= 10;
  }
  return s;
}
int last(uint64_t v) { return v % 10; }
uint64_t suffix(uint64_t v) {
  assert(!(v % 10));
  uint64_t s = 1;
  while (!(v % 10)) {
    v /= 10;
    s *= 10;
  }
  return s;
}
int main() {
  int t;
  std::cin >> t;
  while (t--) {
    uint64_t n, s, nv;
    std::cin >> n;
    std::cin >> s;
    uint64_t c = 0;
    int sn = sum(n);
    nv = n;
    while (sn > s) {
      const int delta = 9 - last(n);
      nv += last(nv) ? (delta + 1) : suffix(nv);
      sn = sum(nv);
    }
    std::cout << nv - n << std::endl;
  }
}

### Prompt

In CPP, your task is to solve 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>
int sum(uint64_t v) {
  int s = 0;
  while (v) {
    s += v % 10;
    v -= v % 10;
    v /= 10;
  }
  return s;
}
int last(uint64_t v) { return v % 10; }
uint64_t suffix(uint64_t v) {
  assert(!(v % 10));
  uint64_t s = 1;
  while (!(v % 10)) {
    v /= 10;
    s *= 10;
  }
  return s;
}
int main() {
  int t;
  std::cin >> t;
  while (t--) {
    uint64_t n, s, nv;
    std::cin >> n;
    std::cin >> s;
    uint64_t c = 0;
    int sn = sum(n);
    nv = n;
    while (sn > s) {
      const int delta = 9 - last(n);
      nv += last(nv) ? (delta + 1) : suffix(nv);
      sn = sum(nv);
    }
    std::cout << nv - n << std::endl;
  }
}

```