#include <iostream>
#include <cassert>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

#define mod 1000000007
#define maxn 100000

int n, k, a[maxn], i, helper[maxn];
double dp[maxn];
priority_queue<pair<double, int> > pq;

int main (int argc, char * const argv[]) {
	scanf("%d %d", &n, &k); // reading the number of streets and k 
	assert(1 <= n && n <= 100000); // input validation
	assert(1 <= k && k <= n); // input validation
	for(i = 0; i < n; i++) {
		scanf("%d", &a[i]); // reading the special number of the i-th street
		assert(1 <= a[i] && a[i] <= 100000); // input validation
	}
	/*
	 then we will use the following observation: a * b < c * d <=> log(a) + log(b) < log(c) + log(d)
	 so we can store minimal logarithms' sum on the way to the i-th street in dp[i]
	 helper[i] is the product modulo 1000000007 itself
	*/
	dp[0] = log(a[0]); // initializing dp for the first street
	helper[0] = a[0]; // initializing helper for the first street
	pq.push(make_pair(-dp[0], 0)); // using priority queue in order to the the optimal "from" street
	for(i = 1; i < n; i++) {
		while (i - pq.top().second > k) pq.pop(); // delete all the streets, from which we're no longer able to reach the current
		dp[i] = dp[pq.top().second] + log(a[i]); // take the optimal previous street to recalculate dp
		helper[i] = (helper[pq.top().second] * 1LL * a[i]) % mod; // and helper
		pq.push(make_pair(-dp[i], i)); // adding this street to the priority queue to be able to recalculate some dp values from it
	}
	printf("%d\n", helper[n - 1]); // outputting an answer
    return 0;
}