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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

#define max(a, b) ((a) < (b) ? (b) : (a))

int main(){
  int Q;
  scanf("%d", &Q);
  
  while(Q--){
    long long A, B;
    scanf("%lld%lld", &A, &B);
    
    long long num = 0;
    
    long long ng = max(A, B);
    long long ok = 0;
    while(ng - ok > 1){
      long long mid = (ok + ng) / 2;
      if(mid < (A * B - 1) / mid) ok = mid;
      else ng = mid;
    }
    
    num += 2 * ok;
    if(ok >= A || ok >= B) num--;
    if(ok + 1 != A && ok + 1 != B && (ok + 1) * (ok + 1) < A * B) num++;
    
    printf("%lld\n", num);
  }
  
  return 0;
}