Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

5 4
1 4 2 3 5

Sample Output 1

4

Four sequences satisfy the condition: (1), (4,2), (1,4,2), and (5).

Sample Input 2

8 4
4 2 4 2 4 2 4 2

Sample Output 2

7

(4,2) is counted four times, and (2,4) is counted three times.

Sample Input 3

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

Sample Output 3

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
const double pi=2*acos(0.0);
double inff=1.0/0.0;
#define ll           long long int
#define inf          1000000000000000008
#define MOD          1000000007
#define pb           push_back
#define sz(x)        (int)x.size()
#define mp           make_pair
#define lbd          lower_bound
#define ubd          upper_bound
#define ipair        pair<ll,ll> 
#define all(v)       v.begin(),v.end()
#define watch(x)     cout<<(#x)<<" is "<<(x)<<endl;
#define rep(a,b,c)   for(a=(int)b;a<(int)c;a++)    
#define bac(a,b,c)   for(a=(int)b;a>=(int)c;a--)
#define se           second
#define f            first
#define ios          ios::sync_with_stdio(false), cout.tie(0), cin.tie(0);
#define mem(a,val)   memset(a,val,sizeof a)
#define endl         "\n"
#define mxn          100002
#define vi           vector<ll>
#define vii          vector<ipair> 
#define si           set<ll> 
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
/*Variables Start*/
ll x,y,n,i,z,j,k,w,u,q,m,t;
vi v;map<ll,ll> m1;
/*Variables End*/
void solve()
{
  cin>>n>>k;
  v.resize(n+1);
  v[0]=0;
  rep(i,1,n+1)
  {
    cin>>v[i];
    v[i]+=v[i-1];
  }
  rep(i,0,n+1)
  {
    if(i>=k)
    {
      x=labs(v[i-k]-i+k)%k;
      m1[x]--;
    }
    x=labs(v[i]-i)%k;
    w+=m1[x];
    m1[x]++;
  }
  cout<<w<<endl;
}
int main() 
{ 
  int TESTS=1;
  ios;  
  // cin>>TESTS;
  while(TESTS--)
  {
    solve();
  }  
}