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: Time Limit Exceeded)

using System;
// using System.Math;
class Program
{
  static void Main(string[] args)
  {
    // スペース区切りの整数の入力
    int q;
    q = int.Parse(Console.ReadLine());

    long[] a = new long[102];
    long[] b = new long[102];
    for (int i = 0; i < q; i++) {
      string[] input = Console.ReadLine().Split(' ');
      a[i] = long.Parse(input[0]);
      b[i] = long.Parse(input[1]);
    }



    for (int i = 0; i < q; i++) {
      long tk = a[i] * b[i];
      long count = 0;
      long max = a[i] * b[i] - 1;
      long min = 1;

      while (true) {
        if (max * min < tk) {
          if (max == min) {
            count += 1;
          } else {
            count += 2;
          }

          if (a[i] == max || a[i] == min)count--;
          if (b[i] == max || b[i] == min)if (count>0)count--;
          max=max*min/(min+1)+1;
          min++;
        } else {
          max--;
        }
        if (max < min)break;
      }

      Console.WriteLine(count);
    }

  }
}