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: Accepted)

#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <numeric> // accumulate(v.begin(), v.end(), 0)                                                         

using namespace std;
#define ll long long

int main(){
  int Q;
  cin >> Q;
  for(int i=0; i<Q; i++){
    ll A, B;
    cin >> A >> B;
    if(A > B){
      ll tmp = A;
      A = B;
      B = tmp;
    }
    if(A == B)
      cout << 2 * A - 2 << endl;
    else if(A + 1 == B)
      cout << 2 * A - 2 << endl;
    else{
      // ll C = floor(sqrt(A * B - 1));                                                                         
      ll C = sqrt(A * B);
      if(C * C == A * B) C--;
      if(C * (C + 1) >= A * B)
        cout << 2 * C - 2 << endl;
      else
        cout << 2 * C - 1 << endl;
    }
  }
  return 0;
}