codeforce 789 C. Functions again (最大字段和)

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C. Functions again

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:

In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x|means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given arraya.

Input

The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.

The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.

Output

Print the only integer — the maximum value of f.

Examples

input

5
1 4 2 3 1

output

3

input

4
1 5 4 7

output

6

Note

In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].

In the second case maximal value of f is reachable only on the whole array.

可以发现[L,R] 区间的f函数值是从L开始相邻两个数的差的绝对值，加减交替所得的结果。所以就是求整个相邻绝对差加减交替数组的最大字段和，

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sum[1]， -sum[2]，+sum[3]，……和-sum[1]，+sum[2]，-sum[3]……两次最大字段和的最大值

代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MAXN 100010
using namespace std;
#define LL __int64
LL sum[MAXN],f[MAXN],a[MAXN];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%I64d",&a[i]);
        if(i>1)
            sum[i-1]=abs(a[i]-a[i-1]);
    }
    int n1=n-1;
    for(int i=2;i<=n;i+=2)
        sum[i]=-sum[i];
    f[0]=0;
    f[1]=sum[1];
    LL maxn=0;
    for(int i=1;i<=n1;i++)
    {
        f[i]=max(sum[i],f[i-1]+sum[i]);
        if(maxn<f[i])
            maxn=f[i];
    }
    for(int i=1;i<=n1;i++)
        sum[i]=-sum[i];
    f[1]=sum[1];
    for(int i=1;i<=n1;i++)
    {
        f[i]=max(sum[i],f[i-1]+sum[i]);
        if(maxn<f[i])
            maxn=f[i];
    }
    printf("%I64d\n",maxn);
    return 0;
}