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 <bits/stdc++.h>

using namespace std;

#define int long long

int q;
int a, b;

bool solve(int v){
    int pos = v / 2;
    for(int i = max(1ll, pos - 10); i <= min(v, pos + 10); i++){
        int diff = v - i;
        int u = 1 + diff;
        int v = i;
        if(v <= b) v--;
        if(u >= a) u++;
        if(u * v >= a * b) return false;
    }
    return true;
}

main()
{
    ios_base::sync_with_stdio(false);
    cin >> q;
    for(int i = 0; i < q; i++){
        cin >> a >> b;
        if(a > b) swap(a, b);
        int lo = 0, hi = 2000000000;
        while(lo != hi){
            int mi = (lo + hi + 1) >> 1;
            if(solve(mi)) lo = mi;
            else hi = mi - 1;
        }
        cout << lo-1 << endl;
    }
}