#include<bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n, k;
	cin>>n>>k;
	vector<ll> a(n+1), pref(n+1);
	for(int i=1; i<=n; ++i) {
		cin>>a[i];
		pref[i]=(pref[i-1]+a[i])%k;
	}
	map<ll, ll> cnt;
	ll ans=0;
	cnt[0]=1;
	if(k==1) {
		cout<<0;
		return 0;
	}
	for(int i=1; i<=n; ++i) {
		if(i>=k) {
			cnt[(pref[i-k]-i%k+k)%k]--;
		}
		ans += cnt[(pref[i]-i%k+k)%k];
		cnt[(pref[i]-i%k+k)%k]++;
	}
	cout<<ans;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

Given are a sequence of N positive integers A_1, A_2, \ldots, A_N, and a positive integer K.

Find the number of non-empty contiguous subsequences in A such that the remainder when dividing the sum of its elements by K is equal to the number of its elements. We consider two subsequences different if they are taken from different positions, even if they are equal sequences.

Constraints

* All values in input are integers.
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N K
A_1 A_2 \cdots A_N


Output

Print the number of subsequences that satisfy the condition.

Examples

Input

5 4
1 4 2 3 5


Output

4


Input

8 4
4 2 4 2 4 2 4 2


Output

7


Input

10 7
14 15 92 65 35 89 79 32 38 46


Output

8

### Response

```cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n, k;
	cin>>n>>k;
	vector<ll> a(n+1), pref(n+1);
	for(int i=1; i<=n; ++i) {
		cin>>a[i];
		pref[i]=(pref[i-1]+a[i])%k;
	}
	map<ll, ll> cnt;
	ll ans=0;
	cnt[0]=1;
	if(k==1) {
		cout<<0;
		return 0;
	}
	for(int i=1; i<=n; ++i) {
		if(i>=k) {
			cnt[(pref[i-k]-i%k+k)%k]--;
		}
		ans += cnt[(pref[i]-i%k+k)%k];
		cnt[(pref[i]-i%k+k)%k]++;
	}
	cout<<ans;
}

```