C

题目链接:http://codeforces.com/problemset/problem/13/C

Little Petya likes to play very much. And most of all he likes to play the following game:

He is given a sequence of N integer numbers. At each step it is allowed to increase the value of any number by 1 or to decrease it by 1. The goal of the game is to make the sequence non-decreasing with the smallest number of steps. Petya is not good at math, so he asks for your help.

The sequence a is called non-decreasing if a₁ ≤ a₂ ≤ ... ≤ a_N holds, where N is the length of the sequence.

Input

The first line of the input contains single integer N (1 ≤ N ≤ 5000) — the length of the initial sequence. The following N lines contain one integer each — elements of the sequence. These numbers do not exceed 10⁹ by absolute value.

Output

Output one integer — minimum number of steps required to achieve the goal.

Examples

Input

5
3 2 -1 2 11

Output

4

Input

5
2 1 1 1 1

Output

1
题目大意：大概的题意就是每次可以对一个数加一或者减一，然后用最小的代价使得原序列变成不下降的序列。
思路：自己最开始的思路是dp[i][j] 第i个数刚好为第j大，并且前i个数满足不递减的关系的最少步数，但是这样想的话要用三重for 所以就超时了 因为我们要枚举从上一个数的所有大小转变过来
这就加了一维了。
正解：dp[i][j]表示前i个树不大于j 并且满足不递减的关系的最少步数
因为是不大于j  所以dp[i][j]可以等于dp[i][j-1]和dp[i-1][j]+abs(a[i]-a[j])  这样就只有二维了
看代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
#include<map>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
#define sc1(a) scanf("%lld",&a)
#define pf1(a) printf("%lld
",a)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const LL INF=1e18;
const ull base=2333;
const int maxn=5e3+5;
const int maxm=1e3+50;
const int maxv=1e6+5;
const int mod=1e9+7;
const int ba=3e5;
LL N;
LL a[maxn],b[maxn];
/**
dp[i][j]表示前i个数最大的数不大于第j大的数 并且前i个数单调不递减的最小步数
*/
LL dp[5][maxn];
int main()
{
    sc1(N);
    for(int i=1;i<=N;i++)
    {
        sc1(a[i]);
        b[i]=a[i];
    }
    sort(b+1,b+1+N);//从小到大
    for(int i=0;i<=1;i++)
    {
        for(int j=0;j<=N;j++) dp[i][j]=INF;
    }
    for(int i=1;i<=N;i++)
    {
        dp[0][i]=0;
    }
    for(int i=1;i<=N;i++)
    {
        int v=i%2;
        int w=(i-1)%2;
        for(int j=0;j<=N;j++) dp[v][j]=INF;
        for(int j=1;j<=N;j++)
        {
            dp[v][j]=min(dp[v][j-1],dp[w][j]+abs(a[i]-b[j]));
        }
    }
    LL ans=INF;
    int v=N%2;
    for(int i=1;i<=N;i++) ans=min(ans,dp[v][i]);
    pf1(ans);
    return 0;
}
/**

*/