D. Plus Minus Permutation

time limit per test

1 second

memory limit per test

256 megabytes

You are given 33 integers — nn, xx, yy. Let's call the score of a permutation†† p1,…,pnp1,…,pn the following value:

(p1⋅x+p2⋅x+…+p⌊nx⌋⋅x)−(p1⋅y+p2⋅y+…+p⌊ny⌋⋅y)(p1⋅x+p2⋅x+…+p⌊nx⌋⋅x)−(p1⋅y+p2⋅y+…+p⌊ny⌋⋅y)

In other words, the score of a permutation is the sum of pipi for all indices ii divisible by xx, minus the sum of pipi for all indices ii divisible by yy.

You need to find the maximum possible score among all permutations of length nn.

For example, if n=7n=7, x=2x=2, y=3y=3, the maximum score is achieved by the permutation [2,6–,1–,7–,5,4––,3][2,6_,1_,7_,5,4__,3] and is equal to (6+7+4)−(1+4)=17−5=12(6+7+4)−(1+4)=17−5=12.

†† A permutation of length nn is an array consisting of nn distinct integers from 11 to nn in any order. For example, [2,3,1,5,4][2,3,1,5,4] is a permutation, but [1,2,2][1,2,2] is not a permutation (the number 22 appears twice in the array) and [1,3,4][1,3,4] is also not a permutation (n=3n=3, but the array contains 44).

Input

The first line of input contains an integer tt (1≤t≤1041≤t≤104) — the number of test cases.

Then follows the description of each test case.

The only line of each test case description contains 33 integers nn, xx, yy (1≤n≤1091≤n≤109, 1≤x,y≤n1≤x,y≤n).

Output

For each test case, output a single integer — the maximum score among all permutations of length nn.

Example

Input

Copy

8

7 2 3

12 6 3

9 1 9

2 2 2

100 20 50

24 4 6

1000000000 5575 25450

4 4 1

Output

Copy

12
-3
44
0
393
87
179179179436104
-6

Note

The first test case is explained in the problem statement above.

In the second test case, one of the optimal permutations will be [12,11,2–,4,8,9––,10,6,1–,5,3,7––][12,11,2_,4,8,9__,10,6,1_,5,3,7__]. The score of this permutation is (9+7)−(2+9+1+7)=−3(9+7)−(2+9+1+7)=−3. It can be shown that a score greater than −3−3 can not be achieved. Note that the answer to the problem can be negative.

In the third test case, the score of the permutation will be (p1+p2+…+p9)−p9(p1+p2+…+p9)−p9. One of the optimal permutations for this case is [9,8,7,6,5,4,3,2,1][9,8,7,6,5,4,3,2,1], and its score is 4444. It can be shown that a score greater than 4444 can not be achieved.

In the fourth test case, x=yx=y, so the score of any permutation will be 00.

解题说明：此题是一道数学题，设z为x与y的最小公倍数，本质上是找到n/x与n/y还有n/z这三个数字是多少，然后这就能确定两个数列的长度分别为n/x-n/z与n/y-n/z。然后就能高斯求和找到1到n中最大的（n/x-n/z）个数相加再减去1到n中最小的（n/y-n/z）个数的和。

#include<stdio.h>
long long int gcd(long long int x, long long int y)
{long long int t;if (y > x){t = y;y = x;x = t;}while (y > 0){x = x % y;t = x;x = y;y = t;}return x;
}
int main()
{long long int t;scanf("%lld", &t);for (long long int i = 0; i < t; i++){long long int n, x, y;long long int a, b;long long int z;scanf("%lld %lld %lld", &n, &x, &y);a = n / x;b = n / y;z = x * y / gcd(x, y);z = n / z;a = a - z;b = b - z;a = n * a - (a - 1) * a / 2;b = b * (b + 1) / 2;printf("%lld\n", a - b);}return 0;
}