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)

#include <bits/stdc++.h>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;

constexpr int Mod = 998244353;
constexpr int mod = 1e9 + 7;
constexpr ll inf = 1LL << 60;

template <typename T> constexpr bool chmax(T &a, const T b) {
    if(a >= b) return false;
    a = b;
    return true;
}
template <typename T> constexpr bool chmin(T &a, const T b) {
    if(a <= b) return false;
    a = b;
    return true;
}

/*-------------------------------------------*/

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    int Q;
    cin >> Q;
    rep(_, Q) {
        ll A, B;
        cin >> A >> B;
        if(A > B) swap(A, B);
        if(A >= B - 1)
            cout << 2 * A - 2 << endl;
        else {
            ll x = sqrt(A * B);
            if(x * x >= A * B) x--;
            cout << 2 * x - (x * (x + 1) < A * B ? 1 : 2) << endl;
        }
    }

    return 0;
}