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: Accepted)

/*
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#        #     _     _     \    \
#        #    (_)   (_)    /    ; 
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*/

#include <bits/stdc++.h>
#define ll long long
#define null NULL
#define int long long

using namespace std;

const int INF = 1e9;

int query(int a, int b) {
	int p = a * b - 1;
	int l = 0, r = INF;
	while (l != r) {
		int mid = (l + r) / 2;
		if ((mid + 1) * (mid + 1) <= p)
			l = mid + 1;
		else
			r = mid;
	}
	int q = l;
	if (a > b)
		swap(a, b);
	if (q * (q + 1) <= p) {
		if (a <= q) {
			return 2 * q - 1;
		}
		else {
			if (q * (q + 2) <= p)
				return 2 * q;
			else
				return 2 * q - 1;
		}
	}
	else {
		return 2 * q - 2;
	}
}

int n;

signed main() {
	cin >> n;
	for (int i = 0; i < n; ++i) {
		int a, b;
		cin >> a >> b;
		cout << query(a, b) << '\n';
	}
}