Problem:
Score : 700 points

Problem Statement

10^{10^{10}} participants, including Takahashi, competed in two programming contests.
In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th.

The score of a participant is the product of his/her ranks in the two contests.

Process the following Q queries:

In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.

Constraints

1 \leq Q \leq 100

1\leq A_i,B_i\leq 10^9(1\leq i\leq Q)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

Q
A_1 B_1
:
A_Q B_Q

Output

For each query, print the maximum possible number of participants whose scores are smaller than Takahashi's.

Sample Input 1

8
1 4
10 5
3 3
4 11
8 9
22 40
8 36
314159265 358979323

Sample Output 1

1
12
4
11
14
57
31
671644785

Let us denote a participant who was ranked x-th in the first contest and y-th in the second contest as (x,y).

In the first query, (2,1) is a possible candidate of a participant whose score is smaller than Takahashi's. There are never two or more participants whose scores are smaller than Takahashi's, so we should print 1.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
  int t;
  cin >> t;
  while(t--){
    int a, b;
    cin >> a >> b;
    if(a > b)swap(a, b);
    ll rnk = 1LL * a * b;
    ll lo = 1, hi = rnk - 1;
    while(lo < hi){
      ll mid = (lo + hi + 1) / 2;
      ll m = mid / 2;
      ll F = mid + (mid >= a ? 1 : 0);
      ll S = mid + (mid >= b ? 1 : 0);
      bool g = 1;
      for(ll x = max(1LL, m - 50);x <= min(m + 50, mid);x++){
        ll L = x;
        ll R = mid - x + 1;
        if(L >= a)
          L++;
        if(R >= b)
          R++;
        if((rnk + R - 1) / R <= L || rnk <= L * R){
          g = 0;
          break;
        }

      }
      if(g)
        lo = mid;
      else
        hi = mid - 1;
    }
    cout << lo << endl;
  }
  return 0;
}