#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T READ() {
  T X = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    X = (X << 1) + (X << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return X;
  return ~(X - 1);
}
int re(long long n) {
  int sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  int T = READ<int>();
  while (T--) {
    long long n = READ<long long>();
    int s = READ<int>();
    long long u = n;
    int now = re(n);
    long long pow2 = 10;
    while (now > s) {
      n /= pow2;
      n++;
      n *= pow2;
      pow2 *= 10;
      now = re(n);
    }
    cout << n - u << endl;
  }
}

### Prompt

Your task is to create a Cpp solution to 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>
using namespace std;
template <typename T>
inline T READ() {
  T X = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    X = (X << 1) + (X << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return X;
  return ~(X - 1);
}
int re(long long n) {
  int sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  int T = READ<int>();
  while (T--) {
    long long n = READ<long long>();
    int s = READ<int>();
    long long u = n;
    int now = re(n);
    long long pow2 = 10;
    while (now > s) {
      n /= pow2;
      n++;
      n *= pow2;
      pow2 *= 10;
      now = re(n);
    }
    cout << n - u << endl;
  }
}

```