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The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn’t have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.

Given a sequence of n integers p1, p2, …, pn. You are to choose k pairs of integers:

[l1, r1], [l2, r2], …, [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < … < lk ≤ rk ≤ n; ri - li + 1 = m), 

in such a way that the value of sum is maximal possible. Help George to cope with the task.

Input

The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, …, pn (0 ≤ pi ≤ 109).

Output

Print an integer in a single line — the maximum possible value of sum.

Examples

Input 5 2 1 1 2 3 4 5 Output 9 Input 7 1 3 2 10 7 18 5 33 0 Output 61 题意：从n个数中选出k段长度为m的连续数组，使得这k段加和最大。 所选取的段数可以是连续的，也可以不是连续的。我们使dp[i][j]代表着从前i个数字中选取j段的最大值。对于dp[i][j]来说，它可以是前i-1个数选取j段，也可以是前i-m段选取j-1段之后，再选取一段。 状态转移方程：dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+cnt);//cnt为从i-m到i的数字和。 代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=5e3+100;ll a[maxx],sum[maxx];ll dp[maxx][maxx];int n,m,k;int main(){
   	scanf("%d%d%d",&n,&m,&k);	for(int i=1;i<=n;i++) scanf("%lld",&a[i]),sum[i]=sum[i-1]+a[i];	int x=m-1;ll cnt;	for(int i=m;i<=n;i++)	{
   		cnt=sum[i]-sum[i-x-1];		for(int j=1;j<=k;j++)		{
   			dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+cnt);		}	}	printf("%lld\n",dp[n][k]);}
   

努力加油a啊，(^o)/~

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