codeforces 895B XK Segments 思维 二分

While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array a and integer x. He should find the number of different ordered pairs of indexes (i, j) such that ai ≤ aj and there are exactly k integers y such that ai ≤ y ≤ aj and y is divisible by x.

In this problem it is meant that pair (i, j) is equal to (j, i) only if i is equal to j. For example pair (1, 2) is not the same as (2, 1).

Input

The first line contains 3 integers n, x, k (1 ≤ n ≤ 105, 1 ≤ x ≤ 109, 0 ≤ k ≤ 109), where n is the size of the array a and x and k are numbers from the statement.

The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.

Output

Print one integer — the answer to the problem.

Example

Input

4 2 1

1 3 5 7

Output

3

Input

4 2 0

5 3 1 7

Output

4

Input

5 3 1

3 3 3 3 3

Output

25

Note

In first sample there are only three suitable pairs of indexes — (1, 2), (2, 3), (3, 4).

In second sample there are four suitable pairs of indexes(1, 1), (2, 2), (3, 3), (4, 4).

In third sample every pair (i, j) is suitable, so the answer is 5 * 5 = 25.

题意:输入n,x,k; 和一个 n 长度的 序列 , 一个二元组的定义如下(i,j)

满足 ai<= y <= aj, 使得y|x的y个数为k个(y为[ai,aj]闭区间的整数)

注意 (i,j) != (j,i) 如果 i!=j的情况下,求出满足这样的二元组的个数

考虑无论从左还是右，低的都会对高的有影响。所以我们不用考虑 ij的位置，把全部的排序，排序后，考虑找到 包含k个x的范围 x*k和 x*k+x-1,同时对于k==0的情况下，都比x小的情况是没办法得出左右边界，

就是数据

2 5 0

4 3 都在0和5的范围内，所以左边界要和a[i] 比较，一定要大于a[i]。为什么是取更大，因为这样不用考虑0

和负数的影响

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[100010];

int main()
{
int n,x,k;
cin>>n>>x>>k;
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
sort(a+1,a+n+1);
ll ans=0;
for(int i=1;i<=n;i++)
{
int tt=(a[i]-1)/x;
tt+=k;
ll l=1ll*tt*x,r=1ll*tt*x+x-1;
l=max(l,a[i]);
ans+=upper_bound(a+1,a+n+1,r)-a-(lower_bound(a+1,a+n+1,l)-a);
}
cout<<ans<<endl;
}

// 3 5 0
// 4 4 4

// 2 5 0
// 3 2