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 <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<int> A(N);
	vector<int> B(N);
	int res = 0;
	for (int i = 0; i < N; i++) {
		cin >> A[i] >> B[i];
	}
	int C;
	int k;
	set<int> X;
	int t = 0;
	for (int i = 0; i < N; i++) {
		C = A[i] * B[i];
		k = sqrt(A[i] * B[i]);
		if (k*k >= A[i] * B[i]) {
			k--;
		}
		t++;
		res = max((int)0,k - 100);
		int a = max((int)0, k - 100);
		int b = (int)1 << 62;
		int bb;
		X.clear();
		//cerr << "ok" << endl;
		while (true) {
			a += 1;
			if (a > k)break;
			if (A[i] == a)continue;
			bb = C / a;
			if (bb*a >= C)bb--;
			bb = min(bb, b - 1);
			if (bb == B[i])bb--;
			X.insert(a);
			b = bb;
			//cerr << a << " " << bb << endl;
			res++;
		}
	
		b = max((int)0, k - 100);
		res += max((int)0, k - 100);
		a = (int)1 << 62;
		int aa;
		while (true) {
			b += 1;
			if (b > k)break;
			if (B[i] == b)continue;
			aa = C / b;
			if (aa*b >= C)aa--;
			aa = min(aa, a - 1);
			if (aa == A[i])aa--;
			if (X.count(aa) == 1)continue;
			a = aa;
			//cerr << aa << " " << b << endl;
			res++;
		}
		if (A[i] <= k - 100) {
			res--;
		}
		if (B[i] <= k - 100) {
			res--;
		}

		cout << res << endl;
	}

}