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.Linq;//リストの使用
using System.Collections.Generic;
class Program
{
static long q = long.Parse(Console.ReadLine());//int.Parseは文字列を整数に変換。
static long[,] vertexes = new long[q,2];//ここに受け取る
static long[] checkNum = new long[2];
	static void Main()
	{
    for(long i = 0; i < q; i++)
    {
      long[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);//１行ごとに受け取る
      vertexes[i,0] = nums[0];
      vertexes[i,1] = nums[1];
    }
    
		for(long i = 0; i < q; i++)
    {
      checkNum[0] = vertexes[i,0];
      checkNum[1] = vertexes[i,1];
      Console.WriteLine(Search(0, checkNum[0]*checkNum[1]));
    }
	}

  static long Search(long min, long max)//二分探索法で最大値を求める
  {
  while (true)
    {
    long mid = min + (max - min) / 2;
    if(Check(mid)) min = mid;//さらに大きくても成り立つかも
    else max = mid;

    if(max - min <= 1)
      {
      if(Check(max)) 
        {
        return max;//最大値で成り立つかの確認を優先
        }
      else
        {
        return min;
        }
    break;
      }  
    }
  }

static bool Check(long testNum)
  {
  if(((testNum+2)/2)*((testNum+3)/2) < checkNum[0]*checkNum[1]) return true;
  return false;
  }
  
}