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>
using namespace std;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)
#define all(a) (a).begin(), (a).end()

int Q;

ll solve(ll a, ll b) {
  ll l = 0, r = 1LL<<31;
  while(r - l > 1) {
    ll m = (l + r) / 2;

    ll res = 0;
    for(ll i = max(m/2-10,1LL); i <= min(m/2+10, m); i++) {
      ll j = i;
      if(a <= i) j++;
      ll k = m + 1 - i;
      if(b <=  m + 1 - i) k++;
      res = max(res, j*k);
    }
    if(res < a*b) {
      l = m;
    } else {
      r = m;
    }
  }
  return l;
}

int main () {
  cin >> Q;

  rep(i,Q) {
    ll A, B;
    cin >> A >> B;

    cout << solve(A, B) << endl;
  }

  return 0;
}