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 <math.h>
using namespace std;

int main() {
	int q;cin>>q;
	long long a[q],b[q];
	for(int i=0;i<q;i++){
		cin>>a[i]>>b[i];
	}

	for(int i=0;i<q;i++){
		long long ans=0;
		long long s=a[i]*b[i];
		long long max=b[i];
		long long min=a[i];
		if(a[i]>b[i]){
			max=a[i];
			min=b[i];
		}
		ans+=2*(min-1);
		long long k=sqrt(s);
		if(max-min>1){
			if(k>min){
				if(k*k==s){
					int l=k-min;
					int r=max-k;
					if(r>l){
						ans=ans+l+l-1;
					}else if(r<l){
						ans=ans+r+r-1;
					}else{
						ans=ans+r+r-2;
					}
				}else{
					if(k*(k+1)<s){
					int l=k-min+1;
					int r=max-k+1;
					if(r>l){
						ans=ans+l+l-1;
					}else if(r<l){
						ans=ans+r+r-1;
					}
					}else{
						ans++;
						int l=k-min;
						int r=max-k;
						if(r>l){
							ans=ans+l+l-1;
						}else if(r<l){
							ans=ans+r+r-1;
						}else{
							ans=ans+r+r-2;
						}

					}
				}
			}
		}

//		long long k=sqrt(s)+1;
//		if(min!=max){
//			while(1){
//				if(k*(k-1)<s){
//					break;
//				}
//				k--;
//			}
//			ans+=k-min;
//		}
		cout<<ans<<endl;

	}
//	for(int i=0;i<q;i++){
//		int ans;
//		int s=a[i]*b[i];
//		int add=0;
//		cout<<"s="<<s<<endl;
//		for(int j=a[i]-1;j>0;j--){
//			if((((s-1)/j+1)-1)>b[i]){
//				add=j;
//				break;
//			}
//		}
//		ans+=add;
//		add=0;
//		for(int j=b[i]-1;j>0;j--){
//			cout<<"j="<<j<<endl;
//			if((((s-1)/j+1)-1)>a[i]){
//				cout<<"a"<<endl;
//				add=j;
//				break;
//			}
//		}
//		ans+=add;
//		cout<<ans<<endl;
//	}
	return 0;
}