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>
#pragma optimize("Ofast")
using namespace std;
using ll = long long;
const int N = 1e5 + 5;
int main() {
  int t;
  cin >> t;
  while (t--) {
    ll n, s;
    cin >> n >> s;
    ll F = n;
    deque<ll> Num, digit;
    while (n > 0) {
      Num.push_front(n % 10);
      digit.push_front(n % 10);
      n /= 10;
    }
    Num.push_front(0);
    digit.push_front(0);
    for (int i = 1; i < Num.size(); ++i) Num[i] += Num[i - 1];
    if (Num[Num.size() - 1] <= s) {
      cout << 0 << '\n';
      continue;
    }
    ll f = n, mul = 1;
    bool oke = false;
    digit[digit.size() - 1] = 0;
    for (int i = Num.size() - 2; i >= 1; --i) {
      if (digit[i] <= 9) {
        ++Num[i];
        digit[i]++;
      } else {
        Num[i] -= 8;
        digit[i] = 0;
      }
      if (Num[i] <= s) {
        for (int j = digit.size() - 1; j >= 1; --j) {
          f += (digit[j] * mul);
          mul *= 10;
        }
        oke = true;
      }
      if (oke) break;
      digit[i] = 0;
    }
    if (!oke) {
      digit[0] = 1;
      for (int j = digit.size() - 1; j >= 0; --j) {
        f += (digit[j] * mul);
        mul *= 10;
      }
    }
    cout << f - F << "\n";
  }
  return 0;
}