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: Runtime Error)

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

int main(){

    int q; cin >> q;
    for(int w = 0; w < q; w++){

    long long A, B; cin >> A >> B;
    if(A > B){
        long long tmp = A;
        A = B;
        B = tmp;
    }
    long long score = A*B;
    //A <= Bが成り立つ

    long long ans = 0;
    long long b; 
    if(A != 1){ 
        b = score / (A - 1) - 1;
        ans += A - 1;
    }else{
        b = B - 1;
    }
    //cout << ans << " " << b << endl; 
    long long a = A + 1;

    while(b >= 0){
        if(b == B){
            b--;
            continue;
        }

        long long temp = score / b;
        int r = score % b;
        if(temp > a || (temp == a) && r){
            ans++;
            //cout << a << " " << b << " " << r << endl;
            if(temp - a > 0) break;
            a = temp + 1;
        }
        b--;
    }

    ans += b - 1;

    cout << ans << endl;
    }
    return 0;
}