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)

//https://blog.csdn.net/hesorchen
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define ll long long
#define lowbit(a) (a & (-a))
#define mod 1000000007
#define INF 0x3f3f3f3f
#define endl "\n"
#define MAX 100010

int main()
{
    ll t;
    cin >> t;
    while (t--)
    {
        ll a, b;
        cin >> a >> b;
        ll m = a * b;
        ll ans = 0;
        for (ll i = 1; i < sqrt(m); i++)
        {
            if (m % i >= 0)
            {
                ll s1 = m / i;
                ll s2 = i;
                if (s1 == a || s1 == b || s2 == a || s2 == b)
                    ans++;
                else
                    ans += 2;
            }
        }
        // if (m % (int)sqrt(m) == 0)
        //     if (a != sqrt(m) && b != sqrt(m))
        //         ans++;
        cout << ans << endl;
    }
    return 0;
}