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;

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> P;
typedef long long ll;

#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define SORT(c) sort((c).begin(),(c).end())
#define REVERSE(c) reverse((c).begin(),(c).end())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int INT_INF = 2147483647;
const long long LL_INF = 1LL<<60;
const long long MOD = 1000000007;

#define CLR(a) memset((a), 0 ,sizeof(a))

#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int main(void){
    ll N,K; cin >> N >> K;
    vector<ll> A(N); REP(i,N) cin >> A[i];
    vector<ll> S1(N+1), S2(N+1); S1[0] = 0; S2[0] = 0;
    REP(i,N){
        S1[i+1] = S1[i] + A[i];
        S2[i+1] = (S1[i+1] - (i+1)) % K;
    }

    map<ll,ll> m;
    REP(i,min(N+1,K)){
        if(m.find(S2[i]) == m.end()) m[S2[i]] = 1;
        else m[S2[i]]++;
    }

    ll ans = 0;
    for(auto p : m){
        ans += p.second * (p.second-1) / 2;
    }

    FOR(i,K,N+1){
        if(m[S2[i-K]]) m[S2[i-K]]--;
        else m.erase(m[S2[i-K]]);
        if(m.find(S2[i]) != m.end()) ans += m[S2[i]]++;
        else m[S2[i]] = 1;
    }
    cout << ans << endl;
    return 0;
}