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)

#define __STDC_FORMAT_MACROS
#define p64 PRId64

#include <bits/stdc++.h>

using namespace std;
using ll = int64_t;

ll A, B;

bool check(ll mid)
{
    ll mx = -1000000001;

    ll mHalf = mid / 2;

    ll x;
    bool flg = false;
    if (mHalf < A)
    {
        flg = true;
        x = mHalf;
    }
    else x = mHalf + 1;

    // mid が 奇数ならば 中央の値
    if (mid % 2 == 1) x++;

    if (x == 0) x++;

    if (x == A) x++;

    ll y;
    if (mHalf < B) y = mHalf + 1;
    else y = mHalf;

    if (y == B && flg) y++;
    else if (y == B) y--;

    mx = max(mx, x * y);

    mx = max(mx, (ll)0);

    // mid が 偶数ならば 中央の 左値 右値
    if (mid % 2 == 0)
    {
        x++;
        if (x == A) x++;

        y--;
        if (y == B) y--;

        mx = max(mx, x * y);

        mx = max(mx, (ll)0);
    }

    return mx < A * B;
}

// Bisection Method
// [left, right]
// to narrow the range
ll BisectionMethod(ll left, ll right)
{
    bool flg = false;

    int times = 0;

    // A * B より小さいという条件を満たさない間は
    // 探索を継続する
    while (1 < abs(right - left) || !flg)
    {
        if (100 < times) break;

        ll mid = (left + right) / 2;

        flg = check(mid);

        if (flg) left = mid;
        else right = mid;

        times++;
    }

    return (left + right) / 2;
}

int main()
{
    ll Q;
    cin >> Q;

    while (Q--)
    {
        cin >> A >> B;
        if (B < A) swap(A, B);
        cout << BisectionMethod(0, 2 * B) << endl;
    }

    return 0;
}