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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

typedef long long ll;

int main() {

	int Q;

	scanf("%d",&Q);

	ll A,B;

	int i,j,x;

	for(j=0;j<Q;j++){
		scanf("%lld %lld",&A,&B);

		if(A>B){
			ll tmp=B; 
			B=A;
			A=tmp;
		}

		ll ans=2ll*(A-1ll);

		int num=0;

		for(i=29;i>=0;i--){
			num+=1<<i;
			if(A+num>B){
				num-=1<<i;
				continue;
			}

			int ng=0;
			for(x=(num/2)-2;x<=(num/2)+2;x++){
				if(x<=0||x>num){continue;}
				if((A+x)*(A+num-x)>=A*B){ng=1;}
			}
			if(ng==1){num-=1<<i;}
		}
		ans+=(ll)num;
		printf("%lld\n",ans);
	}

	return 0;
}