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>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define min(p,q)((p)<(q)?(p):(q))

/*
A≦Bとして一般性を失わない。N=ABとおく。
x[i]*y[i]＜Nとなるものをできるだけ多く作りたい。
z=floor(√(N-1))とする。
x[i]≦z or y[i]≦zが必要。逆にこれを満たす2z個のうち使えないものは
１．z(z+1)≧Nならx[i]=zとy[i]=zは同じ組になるので-1
２．A≦zならx[i]=Aの組は使えないので-1
の2パターンに限る
*/

int main(){
	ll q;
	scanf("%lld",&q);
	while(q--){
		ll a,b;
		scanf("%lld%lld",&a,&b);		
		ll n=a*b;
		//n-1≧2^53になるときに誤差が怖いので
		ll c=sqrt(n-1)-1;
		while((c+1)*(c+1)<=n-1)c++;
		//
		ll ans=2*c;
		if(c*(c+1)>=n)ans--;
		if(min(a,b)<=c)ans--;
		printf("%lld\n",ans);
	}
	return 0;
}