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<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define MOD 1000000007
#define MOD2 998244353
#define pb emplace_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define sz(x) (ll)x.size()
#define F first
#define S second
#define FOR(i,a,b) for(ll i=a;i<=b;++i)
#define ROF(i,a,b) for(ll i=a;i>=b;--i)
#define trace(x) cerr<<#x<<": "<<x<<'\n';
typedef long long ll;
using namespace std;
using namespace __gnu_pbds;
#define T int
#define ordered_set tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>
auto clk=clock();
 
ll mexp(ll a, ll b, ll m){
    ll ans=1;
    a%=m;
    while(b){
        if(b&1) ans=ans*a%m;
        b>>=1;
        a=a*a%m;
    }
    return ans;
}


int main(){
    IOS
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());
    int q;
    cin >> q;
    while(q--){
        ll a, b;
        cin >> a >> b;
        if((a == b) || (abs(a - b) == 1)){
            cout << 2 * (a - 1) << endl;
        }
        else{
            int c = sqrt(a * b);
            if(c * c == a * b){
                c--;
            }
            if(c * (c + 1) >= a* b){
                cout << 2 * c - 2 << endl;
            }
            else{
                cout << 2 * c - 1 << endl;
            }
        }

    }

    cerr<<endl<<endl<<"Time: "<<fixed<<setprecision(12)<<(long double)(clock()-clk)/CLOCKS_PER_SEC<<" ms"<<endl<<endl;
 
    return 0;
}