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)

#define rep(i,a,b) for (int i = (a); i < (b); i++)
#define _USE_MATH_DEFINES
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <functional>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <math.h>
#include <complex>
#include <cassert>
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> p;
typedef complex<double> com;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const ll INF = 1e18;


int main() {
	int n; cin>>n;
	ll ans;
	rep(i,0,n){
		int p1,p2; cin>>p1>>p2;
		if(p1>p2) swap(p1,p2);
		if(p1==p2||p1+1==p2) ans=2*p1-2;
		else{
			int c=sqrt((ll)p1*p2-1);
			if((ll)c*(c+1)>=(ll)p1*p2) ans=2*c-2;
			else ans=2*c-1;
		}
		cout<<ans<<endl;
 	}
	return 0;
}