Codeforces Round #448 (Div. 2)B

B. XK Segments

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

4 2 11 3 5 7

output

3

input

4 2 05 3 1 7

output

4

input

5 3 13 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.

自己真的很菜啊

 这个题想了好久 看了别人的代码才看出来 

找到满足的区间 分别找到与之相邻的下标相减

#include<bits/stdc++.h>using namespace std;#define maxn 1000000+100#define ll long longlong long a[maxn];int main(){   ll n,k,x;   scanf("%lld%lld%lld",&n,&x,&k);   for(ll j=0;j<n;j++){      scanf("%lld",&a[j]);   }   sort(a,a+n);   long long ans=0;   for(ll j=0;j<n;j++){      long long z=a[j]/x;      long long l=(z+k)*x;      long long r=l+x;      if(a[j]%x==0){         l=(z+k-1)*x;         r=l+x;      }      l=max(a[j],l);      //cout<<l<<" "<<r<<" ";      l=lower_bound(a,a+n,l)-a;      r=lower_bound(a,a+n,r)-a;      //cout<<l<<" "<<r<<endl;      ans+=r-l;   }   cout<<ans<<endl;}

#include<bits/stdc++.h>using namespace std;#define maxn 1000000+100#define ll long longlong long a[maxn];ll binary_search1(ll l,ll r,ll x){    while(l<=r){        int mid=(l+r)/2;        if(a[mid]>=x){             r=mid-1;        }        else l=mid+1;    }    return l;}int main(){   ll n,k,x;   scanf("%lld%lld%lld",&n,&x,&k);   for(ll j=0;j<n;j++){      scanf("%lld",&a[j]);   }   sort(a,a+n);   long long ans=0;   for(ll j=0;j<n;j++){      long long z=a[j]%x;     // cout<<a[j]+x-z<<" ";      long long l=a[j]+(x-z)%x+(k-1)*x;      long long r=l+x;      //l=max(a[j],l);      if(!k){         l=a[j];         r=a[j]+(x-z)%x;      }      //cout<<l<<" "<<r<<" ";      l=binary_search1(0,n-1,l);      r=binary_search1(0,n-1,r);      //cout<<l<<" "<<r<<endl;      ans+=r-l;   }   cout<<ans<<endl;}