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 <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#define int long long
using namespace std;

int q;
int a, b;

int solve() {
	int x = (int)sqrt(a * b - 1.0 + 1e-10);
	int ret = 2 * x;
	x = a * b - 1;
	
	int st = 1, ed = 1000000001, mid; //oooxxx, [st, ed)
	while (ed - st >= 2) {
		mid = (st + ed) / 2;
		if (mid <= x / mid) st = mid;
		else ed = mid;
	}
	if (st == x / st) { ret--; }
	
	int a1 = x / a;
	int a2 = x / (a + 1);
	if (a1 - a2 - (a2 < b && b <= a1) > 0) { ret--; } //(a2, a1]からbを消したときに要素が存在するか？
	
	int b1 = x / b;
	int b2 = x / (b + 1);
	if (b1 - b2 - (b2 < a && a <= b1) > 0) { ret--; }
	return ret;
}

signed main() {
	int i;
	
	cin >> q;
	for (i = 0; i < q; i++) {
		cin >> a >> b;
		cout << solve() << endl;
	}
	return 0;
}