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: Time Limit Exceeded)

#include<stdio.h>

long long int nibutan(long long int left, long long int right, long long int num) {
	long long int memo = (left + right) / 2;
	if (memo*(memo + 1) < num && (memo + 2)*(memo + 1) >= num) { return memo; }
	if (memo*(memo + 1) < num && (memo + 2)*(memo + 1) < num) { return nibutan(memo, right, num); }
	if (memo*(memo + 1) >= num && (memo + 2)*(memo + 1) >= num) { return nibutan(left, memo, num); }
}
long long int nibutan2(long long int left, long long int right, long long int num) {
	long long int memo = (left + right) / 2;
	if (memo*memo < num && (memo + 1)*(memo + 1) >= num) { return memo; }
	if (memo*memo < num && (memo + 1)*(memo + 1) < num) { return nibutan2(memo, right, num); }
	if (memo*memo >= num && (memo + 1)*(memo + 1) >= num) { return nibutan2(left, memo, num); }
}

int main() {

	long long int n, memo, r;
	long long int a[105], b[105];
	scanf("%lld", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lld%lld", &a[i], &b[i]);
		memo = a[i] * b[i];


		if (a[i] == b[i]) {
			printf("%lld\n",2*(a[i]-1));
		}
		else if (a[i] >= b[i] && nibutan(1, a[i], memo) >= nibutan2(1, a[i], memo)) {


			printf("%lld\n", 2 * nibutan(1, a[i], memo) - 1);
		}
		else if (a[i] < b[i] && nibutan(1, b[i], memo) >= nibutan2(1, b[i], memo)) {
			printf("%lld\n", 2 * nibutan(1, b[i], memo) - 1);

		}
		else if (a[i] >= b[i] && nibutan(1, a[i], memo) < nibutan2(1, a[i], memo)) {
			printf("%lld\n", 2 * nibutan(1, a[i], memo));
		}
		else if (a[i] < b[i] && nibutan(1, b[i], memo) < nibutan2(1, b[i], memo)) {
			printf("%lld\n", 2 * nibutan(1, b[i], memo));
		}
	}

}