CodeForces - 1066B Heaters （水题，模拟+二分）

Vova's house is an array consisting of nn elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The ii-th element of the array is 11 if there is a heater in the position ii, otherwise the ii-th element of the array is 00.

Each heater has a value rr (rr is the same for all heaters). This value means that the heater at the position pospos can warm up all the elements in range [pos−r+1;pos+r−1][pos−r+1;pos+r−1].

Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.

Vova's target is to warm up the whole house (all the elements of the array), i.e. ifn=6n=6, r=2r=2 and heaters are at positions 22 and 55, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 33 elements will be warmed up by the first heater and the last 33 elements will be warmed up by the second heater).

Initially, all the heaters are off.

But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.

Your task is to find this number of heaters or say that it is impossible to warm up the whole house.

Input

The first line of the input contains two integers nn and rr (1≤n,r≤10001≤n,r≤1000) — the number of elements in the array and the value of heaters.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤10≤ai≤1) — the Vova's house description.

Output

Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.

Examples

Input

6 2
0 1 1 0 0 1

Output

Input

5 3
1 0 0 0 1

Output

Input

5 10
0 0 0 0 0

Output

-1

Input

10 3
0 0 1 1 0 1 0 0 0 1

Output

Note

In the first example the heater at the position 22 warms up elements [1;3][1;3], the heater at the position 33 warms up elements [2,4][2,4] and the heater at the position 66 warms up elements [5;6][5;6] so the answer is 33.

In the second example the heater at the position 11 warms up elements [1;3][1;3] and the heater at the position 55 warms up elements [3;5][3;5] so the answer is 22.

In the third example there are no heaters so the answer is -1.

In the fourth example the heater at the position 33 warms up elements [1;5][1;5], the heater at the position 66 warms up elements [4;8][4;8] and the heater at the position 1010warms up elements [8;10][8;10] so the answer is 33.

数值为1的地方可以点燃它并影响【i-r+1,i+r-1】的区域，问最少点燃几个地方就能影响所有位置。不能的话就输出-1；

预处理后使用二分查找，每次查找最大的能够影响目前位置的点在哪里。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int n, m[maxn], t;
int sot[maxn], cnt;
int main() {
	while (~scanf("%d%d", &n, &t)) {
		cnt = 1;
		for (int i = 1; i <= n; i++) {
			int tmp;
			scanf("%d", &tmp);
			if (tmp){
				sot[cnt++] = i - t + 1;
			}
		}
		int now = 1, spot = 1, sum = 0;
		while (now <= n) {
			int pos = upper_bound(sot + 1, sot + cnt, now) - sot;
		//	cout << pos <<" "<<now<< endl;
			if (pos == 1 || sot[pos - 1] + 2*t - 2 < now) {
				spot = 0; break;
			}
			sum++;
			now = sot[pos - 1] + 2 * t - 1;
		}
		if (spot)
			cout << sum << endl;
		else
			cout << -1 << endl;
	}
}