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 <bits/stdc++.h>

using namespace std;

using ll = long long;
using ull = unsigned long long;

int main(int argc, char const *argv[]) {
	int q;
	std::cin >> q;
	std::vector<std::vector<double>> v(q,std::vector<double>(2));
	double a,b;
	for (size_t i = 0; i < q; i++) {
		std::cin >> v[i][0] >> v[i][1];
	}
	for (size_t i = 0; i < q; i++) {
		a = v[i][0];
		b = v[i][1];
		if (b < a) {
			swap(a,b);
		}
		int absqrt = (int)(sqrt(a*b-1.0));
		if (absqrt == 0) {
			std::cout << 0 << '\n';
			continue;
		}

		if (a == b) {
			std::cout << 2 * a - 2 << '\n';
		}
		else if (absqrt * (absqrt + 1) >= a * b) {
			std::cout << absqrt * 2 - 2 << '\n';
		}
		else {
			std::cout << absqrt * 2 - 1 << '\n';
		}


	}

	// これ、高橋君以下の参加者だ……
	/* この問題は多分平方根が求まれば勝ち
	 * 平方根の整数部分をXとすると
	 * X * (X+1) < A * B
	 * X * (X+1) > A * B
	 * の2パターンで挙動が変わる
	 * X * (X+1) < A * Bのときは
	 * X*2 - 1
	 * X * (X+1) > A * Bのときは
	 * (X-1)*2 + 1 - 1 = X*2 - 2
	 * で求まる。
	 */

	// 知りたいのは高橋君より小さい参加者

	return 0;
}