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: Accepted)

#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll q;
    cin>>q;
    ll a,b;
    vector<ll> ans;

    for(int i=0;i<q;i++){
    	cin>>a>>b;
    	if(a==b){
    		ans.pb(2*(a-1));
    		continue;
    	}
    	if(b>a)swap(a,b);
    	ll ma=1e9+2,mi=0,mid=(ma+mi)/2;
    while(mi!=mid){
    	if(mid*mid>a*b){
    		ma=mid;
    	}
    	else{
    		mi=mid;
    	}
    	mid=(ma+mi)/2;
    }
    if(mi*mi==a*b){
    	ans.pb(mi*2-3);
    }
    else if(a-b==1){
    	ans.pb(mi*2-2);
    }
    else if(mid*(mid+1)<a*b)ans.pb((mi*2-1)); 
    else ans.pb((mi*2-2));
    
    }
    
    for(int i=0;i<q;i++){
    	cout << ans[i]<<endl;
    }
    return 0;
}