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 <iostream>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <stack>
#include <list>
#include <utility>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
#include <climits>
#include <map>
#include <unordered_map>
#include <string.h>
#include<iomanip>
#include <iterator>
#include <locale>

#define all(a) a.begin(),a.end()
#define VV(T) vector<vector< T > >
#define rep(i,N) for(int i = 0; i <N; i++)
#define repsg(i,s,g) for(int i = s; i <g; i++)
#define name(x) #x
using namespace std;

const int INF = INT_MAX;
const long long INFL = 1LL<<40;
const double PI = atan(1)*4;
//const int MOD = pow(10,9)+7;
int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};

typedef long long ll,LL;
typedef vector<int> vint;
typedef vector<vector<int> > vvint;
typedef vector<long long> vll, vLL;
typedef vector<vector<long long> > vvll, vvLL;
typedef pair<int,int> pii;
typedef pair<LL, LL> pLL;


// vector入力
template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
    for(T& x: vec) is >> x;
    return is;
}
// vector出力
template<typename T>
ostream& operator << (ostream& os, vector<T>& vec){
    for(int i=0; i<vec.size(); i++){
        os << vec[i] << ( i+1 == vec.size() ? "" : " " );
    }
    return os;
}


int main() {
    int q;cin>>q;

    rep(i,q){
        ll a,b;cin>>a>>b;
        
        if(a<b)swap(a,b);
        
        if(a==b){
            cout<<(a)*2-2<<endl;
        }else{
            double v = a * b;
            ll c = sqrt(v);
            
            if(c*(c+1)<a*b){
                cout<<2*c-1<<endl;
            }else if(c*c==a*b){
                cout<<2*c-3<<endl;
            }else{
                cout<<2*c-2<<endl;
            }
            
            
        }
        
    }
    
    
    return 0;
}