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 <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <numeric>
#include <cmath>
#include <cassert>

using namespace std;

typedef long long int ll;
typedef pair<int,int> P;

#define rep(i,n) for(int i=0;i<(n);++i)
#define all(x) x.begin(),x.end()

const ll mod = 1e9+7;
const ll INF = 1e9;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

int main()
{
	int q;
	cin>>q;
	while(q--){
		ll a,b;
		cin>>a>>b;
		if(a>b){
			ll tmp = b;
			b=a;
			a=tmp;
		}
		ll ans = 0;

		if(a==b) ans=2*a-1;
		else if(a+1==b) ans=2*a-2;
		else{

			ll c = floor(sqrt(a*b));
			if(c*c==a*b)c--;
			if(c*(c+1)<a*b) ans = 2*c-1;
			else ans = 2*c-2;
		}
		cout<<ans<<endl;
	}
	return 0;
}