//整个机房都是同一位神牛的膜拜者，或者两个神牛的膜拜者人数差不超过M。
#include <cstdio>
using namespace std;

int a[2501]={0},sum[3][2501]={{0}},f[2501][2501]={{0}};

int absint(int a)
{
	if (a<0)
		return(-a);
	else
		return(a);
}

int main(void)
{
	freopen("orz.in","r",stdin);
	freopen("orz.out","w",stdout);
	int i,j,k,n,m,temp;
	scanf("%d %d\n",&n,&m);
	for (i=1;i<=n;i++)
	{
		scanf("%d\n",&a[i]);
		sum[a[i]][i]=sum[a[i]][i-1]+1;
		sum[3-a[i]][i]=sum[3-a[i]][i-1];
	}
	for (i=1;i<=n;i++)
		f[i][1]=1;
	for (j=2;j<=n;j++)
		for (i=1;i<=n-j+1;i++)
		{
			if (sum[1][i+j-1]-sum[1][i-1]==0||sum[2][i+j-1]-sum[2][i-1]==0||absint((sum[1][i+j-1]-sum[1][i-1])-(sum[2][i+j-1]-sum[2][i-1]))<=m)
			{
				f[i][j]=1;
				continue;
			}
			else
				f[i][j]=32767;
			for (k=i;k<=i+j-2;k++)
			{
				temp=f[i][k-i+1]+f[k+1][i+j-1-k];
				if (temp<f[i][j])
					f[i][j]=temp;
			}
		}
	printf("%d\n",f[1][n]);
	fclose(stdin);
	fclose(stdout);
	return(0);
}