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<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
using namespace std;
#define N 
#define ll long long
#define db double
#define eps 1e-9
#define dbg1(x) cout<<#x<<"="<<x<<" "
#define dbg2(x) cout<<#x<<"="<<x<<"\n"
int T;ll a,b,ans;
//db cal(ll x) {return (-(ll)b*x-1)*1./(a+x);}
db cal(ll x) {return -b*1.0/(a+x)*x-1./(a+x);}
int main()
{
//	freopen("1.in","r",stdin);
//	freopen("1.out","w",stdout);
	for(scanf("%d",&T);T--;)
	{
		scanf("%lld%lld",&a,&b),ans=0;
		if(a<b) swap(a,b);
		if(a==1&&b==1) {puts("0");continue;}
		/*if(b>=20000000)
		{
			ll dd=sqrt(1ll*a*b),ans=dd*2-1;
			if(dd*dd==1ll*a*b)
			{
				if(a==b)ans--;
				else ans-=2;
			}
			else if(dd*(dd+1)>=1ll*a*b)ans--;
			cout<<ans<<endl;
			continue;
		}
		
		for(ll x=-a+1,y,tmp;;x++)
		{
			tmp=floor(((ll)-b*x-1)*1./(a+x));
			dbg1(x),dbg2(tmp);
			if(tmp==-b) break;
			if(tmp)
			{
				if(tmp^last) ans++,last=tmp;
				else
				{
					ans+=tmp+b-1;
					if(tmp>0) ans--;
					break;
				}
			}
		}
		dbg2(ans);*/
		ll l=-a+1,r=2e9,mid;
		for(;l<r;)
		{
			mid=l+r>>1;
//			dbg2(mid);
//			dbg1(cal(mid+1)),dbg2(cal(mid));
			cal(mid)-cal(mid+1)-1<-eps?r=mid:l=mid+1;
		}
//		dbg2(l);
		ll tmp1=floor((-(ll)b*l-1)*1./(a+l));
		ans=l+a+tmp1+b-1;
		if(l==-a+1)
		{
			if(tmp1>=0) ans--;
			cout<<ans<<"\n";
			continue;
		}
		ll tmp2=floor((-(ll)b*(l-1)-1)*1./(a+l-1));
//		dbg1(l-1),dbg1(cal(l-1)),dbg2(tmp2);
//		dbg1(l),dbg1(cal(l)),dbg2(tmp1);
//		dbg1(tmp1),dbg2(tmp2);
		if(tmp1>=0||tmp2<=0) ans--;
//		printf("%lld\n",ans);
		cout<<ans<<"\n";
	}
	return 0;
}