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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdint.h> // int64_t
#include <math.h>

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))

int get_int(void) {
  int num;
  scanf("%d", &num);
  return num;
}

int get_int2(int *a1, int *a2) {
  scanf("%d %d", a1, a2);
  return 0;
}

void swap(int *a1, int *a2) {
    int tmp = *a1;
    *a1 = *a2;
    *a2 = tmp;
}

#define QUERY_MAX 100

int main(void) {
    int queries = get_int();

    int i;
    int64_t ans[QUERY_MAX];
    int aidx = 0;
    for(i = 0; i < queries; i++) {
        int a1, a2;
        int64_t res;
        get_int2(&a1, &a2);
        if(a1 > a2) swap(&a1, &a2);
        if(a1 == a2) {
            res = (int64_t)(a1-1)*2; // exclude a1-a2
        } else if (a1+1 == a2) {
            res = (int64_t)(a1-1)*2;
        } else {
            int64_t sum = (int64_t)a1*a2;
            int64_t power_2 = sqrt(-0.5 + sum);
            if(power_2*(power_2+1) >= sum) {
                res = (power_2-1)*2;
            } else {
                res = (power_2)*2-1;
            }
            if(sum <= 3) {
                res = 1;
            }
        }
        ans[aidx++] = res;
    }

    for(i = 0; i < aidx; i++) {
        printf("%lld\n", ans[i]);
    }
    return 0;
}