A. The Beatles

Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n⋅k cities. The cities are numerated from 11to n⋅kn⋅k, the distance between the neighboring cities is exactly 11 km.

Sergey does not like beetles, he loves burgers. Fortunately for him, there are nn fast food restaurants on the circle, they are located in the 11-st, the (k+1)(k+1)-st, the (2k+1)(2k+1)-st, and so on, the ((n−1)k+1)((n−1)k+1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is kk km.

Sergey began his journey at some city ss and traveled along the circle, making stops at cities each ll km (l>0l>0), until he stopped in ss once again. Sergey then forgot numbers ss and ll, but he remembers that the distance from the city ss to the nearest fast food restaurant was aakm, and the distance from the city he stopped at after traveling the first ll km from ss to the nearest fast food restaurant was bb km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.

Now Sergey is interested in two integers. The first integer xx is the minimum number of stops (excluding the first) Sergey could have done before returning to ss. The second integer yy is the maximum number of stops (excluding the first) Sergey could have done before returning to ss.

Input

The first line contains two integers nn and kk (1≤n,k≤1000001≤n,k≤100000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.

The second line contains two integers aa and bb (0≤a,b≤k20≤a,b≤k2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.

Output

Print the two integers xx and yy.

Examples

input

Copy

2 3
1 1

output

Copy

1 6

input

Copy

3 2
0 0

output

Copy

1 3

input

Copy

1 10
5 3

output

Copy

5 5

Note

In the first example the restaurants are located in the cities 11 and 44, the initial city ss could be 22, 33, 55, or 66. The next city Sergey stopped at could also be at cities 2,3,5,62,3,5,6. Let's loop through all possible combinations of these cities. If both ss and the city of the first stop are at the city 22 (for example, l=6l=6), then Sergey is at ss after the first stop already, so x=1x=1. In other pairs Sergey needs 1,2,31,2,3, or 66 stops to return to ss, so y=6y=6.

In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so ll is 22, 44, or 66. Thus x=1x=1, y=3y=3.

In the third example there is only one restaurant, so the possible locations of ss and the first stop are: (6,8)(6,8) and (6,4)(6,4). For the first option l=2l=2, for the second l=8l=8. In both cases Sergey needs x=y=5x=y=5 stops to go to ss.

题意：有n*k个城市在（1，k+1,(i-1)*k+1）处有n个餐馆，已知开始的出发的城市距离一个餐馆有akm，第一个到的城市距离最近的餐馆有bkm问出发到回来出发的城市最少多少步，最多多少步。

思路：枚举长度(k*x+b），然后以1为起点，分两个方向找到的第一步到的起点。注意（这里只要枚举一个起点是因为枚举的是出发点和第一个点的长度然后求gcd找出最多一步走多少，最少走多少，然后除一下就得出答案）。

代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long longusing namespace std;//LL a[maxn];
//LL sum[maxn];
LL gcd(LL x,LL y)
{if(y==0)return x;elsereturn gcd(y,x%y);
}
int main()
{LL n,k,a,b;scanf("%lld%lld",&n,&k);scanf("%lld%lld",&a,&b);LL start=1+a;LL mx=1,mi=n*k;LL sum=n*k;for(int i=1;i<=n;i++){LL en=(i-1)*k+b+1;mx=max(mx,gcd(sum,sum+(en-start)%sum));mi=min(mi,gcd(sum,sum+(en-start)%sum));en=(i-1)*k+1-b;mx=max(mx,gcd(sum,sum+(en-start)%sum));mi=min(mi,gcd(sum,sum+(en-start)%sum));}printf("%lld %lld\n",sum/mx,sum/mi);return 0;
}

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