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)

#include <iostream>
#include <cmath>
using namespace std;
// class quad{
// public:
	// double high;
	// double low;
// };

// int main(){
	// int Q;
	// long long A,B;
	// long long AB,tmp1,tmp2;
	// quad qAB;
	// long long C;
	// cin >> Q;
	// for(int i=0; i<Q; i++){
		// cin >> A >> B;
		// AB = A*B;
		// qAB.high = AB;
		// tmp1 = qAB.high;
		// tmp2 =  AB - tmp1;
		// qAB.low = tmp2;
		// cout << AB << endl;
		// cout << qAB.high <<endl;
		// cout << qAB.low << endl;
	// }

	// return 0;
// }
int main(){
	int Q;
	long long A,B;
	double AB;
	long long C;
	cin >> Q;
	for(int i=0; i<Q; i++){
		cin >> A >> B;
		AB = A*B-1;
		C=long(sqrt(AB));
		if(C<A && C <B){
			cout << C*2-1 << endl;
		}else{
			cout << C*2-2 << endl;
		}
	}

	return 0;
}