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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def LS(): return list(map(list, input().split()))
def S(): return list(input().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')

class BinaryIndexedTree:
    # http://hos.ac/slides/20140319_bit.pdf
    def __init__(self, size):
        """
        :param int size:
        """
        self.bit = [0 for _ in range(size)]
        self.size = size

    def add(self, i, w):
        """
        i番目にwを加える
        :param int i:
        :param int w:
        :return:
        """
        x = i + 1
        while x <= self.size:
            self.bit[x - 1] += w
            x += x & -x
        return

    def sum(self, i):
        """
        [0,i]の合計
        :param int i:
        :return:
        """
        res = 0
        x = i + 1
        while x > 0:
            res += self.bit[x - 1]
            x -= x & -x
        return res

    def search(self, x):
        """
        二分探索。和がx以上となる最小のインデックス(>= 1)を返す
        :param int x:
        :return :
        """
        i = 1
        s = 0
        step = 1 << (self.size.bit_length() - 1)
        while step:
            if i + step <= self.size and s + self.bit[i + step - 1] < x:
                i += step
                s += self.bit[i - 1]
            step >>= 1
        return i

    def __len__(self):
        return self.size
#solve
def solve():
    n, k = LI()
    a = LI_()
    ab = a[::1]
    a[0] = a[0] % k
    for i in range(n - 1):
        a[i + 1] = (a[i] + a[i + 1]) % k
    print(a)
    Bit = BinaryIndexedTree(k + 1)
    ans = 0
    d = deque([0])
    Bit.add(0, 1)
    for x, i in enumerate(a):
        d.append(i)
        if len(d) == k+1:
            x = d.popleft()
            Bit.add(x, -1)
        ans += Bit.sum(i) - Bit.sum(i - 1)
        Bit.add(i, 1)
    print(ans)
    return


#main
if __name__ == '__main__':
    solve()