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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using static System.Console;

class Program
{
    static void Main()
    {
        var Q = int.Parse(ReadLine());
        var A = new long[Q];
        var B = new long[Q];
        for (int i = 0; i < Q; i++)
        {
            var AB = ReadLine().Split().Select(long.Parse).ToArray();
            A[i] = AB[0];
            B[i] = AB[1];
        }
        var ans = new long[Q];

        for (int i = 0; i < Q; i++)
        {
            var a = A[i];
            var b = B[i];
            var ua = new HashSet<long>();
            var ub = new HashSet<long>();
            for (int j = 1; j <= a; j++)
            {
                 if(ua.Count==a)break;
                for (long k = a * b / j; k > 0; k -= j)
                {
                    if(ub.Count==b)break;
                    if (!ub.Contains(k) && j != a && k != b)
                    {

                        ua.Add(j);
                        ub.Add(i);
                        break;
                    }
                }
            }
            ans[i]=ua.Count();
        }
        for (int i = 0; i < Q; i++)
        {
            WriteLine(ans[i]);
        }
    }
}