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<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;

typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
typedef long double ld;
const ll mod = 1000000007;
const ll INF = mod * mod;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<int, int> P;
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
typedef vector<int> vec;
typedef vector<string> svec;
#define fr first
#define sc second
#define all(c) c.begin(),c.end()
#define pb push_back
//#define int long long

void solve() {
    int q; cin >> q;
    rep(i, q) {
        ll a, b; cin >> a >> b;
        if(a == b) {
            cout << 2 * (a - 1) << endl;
            continue;
        }
        ll lim = a * b - 1;
        ll ok = 0, ng = 1LL << 32;
        while(ng - ok > 1) {
            ll mid = (ok + ng) / 2;
            ll mul = (mid / 2) * (mid - mid / 2);
            if(mul > lim) ng = mid;
            else ok = mid;
        }
        cout << ok << endl;
    }
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	//init();
	solve();
	//cout << "finish" << endl;
    return 0;
}