#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll  n,c;
ll cun[200002];
ll cun1[200002];
int main()
{
    scanf("%lld%lld",&n,&c);
    for(register int i=1;i<=n;i++)
    {
        scanf("%lld",&cun[i]);
        cun1[cun[i]]++;
        cun[i]-=c;
    }
    long long da=0;
    for(register int i=1;i<=n;i++)
    {
        da+=cun1[cun[i]];
    }
    cout<<da;
    return 0;
}
  /*
  以下题解，用的map
      #include <iostream>
    #include <map>
    using namespace std;
    typedef long long LL;
    LL a[200001];
    map<LL,LL> m;//建立一个数字到出现次数的映射 map<num,times>
    //A-B=C --> A-C=B
    int main() {
        int n;
        LL c;
        LL ans=0;
        cin >> n >> c;
        for(int i=1;i<=n;i++) {
            cin >> a[i];
            m[a[i]]++;
            a[i]-=c;    
        } 
        for(int i=1;i<=n;i++) ans+=m[a[i]];
        cout << ans << endl;
        return 0;
}*/