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

int n;
ll S[201010];
ll a[201010], K;
map<ll, ll> mp;
void solve() {
	scanf("%d%lld", &n, &K);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", a + i);
		S[i] = (S[i - 1] + a[i] - 1) % K;
		//sum-n%K=0;
	}
	ll ans = 0;
	for (int i = 0; i <= n; ++i) {
		ans += mp[S[i]];
		++mp[S[i]];
		if (i >= K - 1)--mp[S[i - K + 1]];
	}
	printf("%lld\n", ans);
}
int main() {
	solve();

	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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;
typedef long long ll;

int n;
ll S[201010];
ll a[201010], K;
map<ll, ll> mp;
void solve() {
	scanf("%d%lld", &n, &K);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", a + i);
		S[i] = (S[i - 1] + a[i] - 1) % K;
		//sum-n%K=0;
	}
	ll ans = 0;
	for (int i = 0; i <= n; ++i) {
		ans += mp[S[i]];
		++mp[S[i]];
		if (i >= K - 1)--mp[S[i - K + 1]];
	}
	printf("%lld\n", ans);
}
int main() {
	solve();

	return 0;
}
```