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

signed main() {
	ll N, K;
	scanf("%lld%lld", &N, &K);
	vector<ll> A(N, 0);
	for(ll i=0; i<N; i++){
		scanf("%lld", &A[i]);
	}

	vector<ll> S(N+1,0);
	for(ll i=0; i<N; i++){
		S[i+1]=S[i]+A[i];
	}

	ll ans = 0;
	map<ll,ll> m;
	for(ll i=0; i<N+1; i++){
		ll j=i-K;
		if(j>=0){
			ll s=(S[j]-j)%K;
			m[s]--;
		}
		ll t=(S[i]-i)%K;
		if(m[t]>0){
			ans+=m[t];
		}
		m[t]++;
	}

	printf("%lld\n", ans);
	return 0;
}

### 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>
using namespace std;
using ll = long long;

signed main() {
	ll N, K;
	scanf("%lld%lld", &N, &K);
	vector<ll> A(N, 0);
	for(ll i=0; i<N; i++){
		scanf("%lld", &A[i]);
	}

	vector<ll> S(N+1,0);
	for(ll i=0; i<N; i++){
		S[i+1]=S[i]+A[i];
	}

	ll ans = 0;
	map<ll,ll> m;
	for(ll i=0; i<N+1; i++){
		ll j=i-K;
		if(j>=0){
			ll s=(S[j]-j)%K;
			m[s]--;
		}
		ll t=(S[i]-i)%K;
		if(m[t]>0){
			ans+=m[t];
		}
		m[t]++;
	}

	printf("%lld\n", ans);
	return 0;
}

```