Preparing Olympiad CodeForces - 550B

You have n problems. You have estimated the difficulty of the i-th one as integer c_i. Now you want to prepare a problemset for a contest, using some of the problems you've made.

A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.

Find the number of ways to choose a problemset for the contest.

Input

The first line contains four integers n, l, r, x (1 ≤ n ≤ 15, 1 ≤ l ≤ r ≤ 10⁹, 1 ≤ x ≤ 10⁶) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.

The second line contains n integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 10⁶) — the difficulty of each problem.

Output

Print the number of ways to choose a suitable problemset for the contest.

Example

Input

3 5 6 1
1 2 3

Output

2

Input

4 40 50 10
10 20 30 25

Output

2

Input

5 25 35 10
10 10 20 10 20

Output

6

题目大意：

从给定的几个数中选出满足大于等于l，小于等于r，且最大于最小的差不能小于x

思路：

这道题就是一个dfs嘛，深度优先的搜索，从第一个开始去搜，搜到符合大于等于l，小于等于r的就加一，其余的就不符合，记得要记录下最大值和最小值，如果排个序的话会简单的点，时间也会用的少点。

代码；

#include<cstdio>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int a[20],l,r,y,n,ans,mn,mx;
void dfs(int x,int sum)
{
    /*printf("x=%d\n",x);
    printf("mn=%d mx=%d\n",mn,mx);*/
    if(sum>=l&&sum<=r&&a[x]-mn>=y)
        ans++;
    for(int i=x+1;i<n;i++)
    {
        if(sum+a[i]<=r)
        {
            /*printf("sum+a[%d]=%d\n",i,sum+a[i]);
            printf("mn=%d mx=%d\n",mn,mx);
            printf("-----------\n");*/

            dfs(i,sum+a[i]);
        }
    }
}
int main()
{
    while(~scanf("%d%d%d%d",&n,&l,&r,&y))
    {
        ans=0;
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        for(int i=0;i<n;i++)
        {
            mn=a[i];
            dfs(i,a[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}