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 <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>

using namespace std;

int main()
{
	int Q;
	long long int a, b1, b, s, ans;
	scanf("%d", &Q);
	for(int i=0; i<Q; i++){
		scanf("%lld %lld", &a, &b);
		if(a<b){
			b1=b;
			b=a;
			a=b1;
		}
		long long int s1=b-1, s2=a+1;
		while(s1!=s2){
			long long int smid=(s1+s2+1)/2;
			if(smid*smid<=a*b){
				s1=smid;
			}else{
				s2=smid-1;
			}
		}
		s=s1;
	

	
		if(b==1){
			ans=0;
		}else if(a*b%(b-1)==0){
			if(a*b/(b-1)-1==a){
				ans=b-2;
			}else{
				ans=b-1;
			}
		}else{
			if(a*b/(b-1)==a){
				ans=b-2;
			}else{
				ans=b-1;
			}
		}
		ans=ans+s-1;
		long long int s0,s3;
		if(a*b%s==0){
			s0=a*b/s-1;
		}else{
			s0=a*b/s;
		}
		if(s0>b){
			ans++;
		}
		if(s0>b){
			if(a*b%(s+1)==0){
				s3=a*b/(s+1)-1;
			}else{
				s3=a*b/(s+1);
			}
			ans=ans+s3-b;
		}


		printf("%lld\n", ans);
	}
	return 0;

}