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 long long INF = 2e18;
vector<long long> v[10005];
vector<long long> color(10005);
vector<long long> col(10005);
void dfs(long long node, long long pnode, long long colour) {
  color[node] = colour;
  for (long long i = 0; i < v[node].size(); i++) {
    if (v[node][i] != pnode && v[node][i] > node) {
      dfs(v[node][i], pnode, colour);
    }
  }
}
long long sudig(long long n) {
  string s = to_string(n);
  long long ans = 0;
  for (long long i = 0; i < s.length(); i++) {
    ans += s[i] - '0';
  }
  return ans;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    long long s;
    cin >> n >> s;
    long long a = sudig(n);
    if (a <= s) {
      cout << 0 << endl;
      continue;
    }
    long long dig = to_string(n).size();
    long long sum = 0;
    string str = to_string(n);
    string str1 = "";
    long long i = dig - 1;
    long long count = 0;
    while (sum < a + 1 - s) {
      sum += str[i] - '0';
      count++;
      str1 += str[i];
      i--;
    }
    reverse(str1.begin(), str1.end());
    long long ans = (long long)pow(10, count) - stoll(str1);
    cout << ans << endl;
  }
}