Functions again

                                                   Functions again  

原题链接 http://codeforces.com/problemset/problem/789/C

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 array a.

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.

给出一个数组，数组符合，从数组中选出一段区间，使函数得到最大值输出即可。

开始没有多想他的关系，后来仔细观察了下函数关系式才知道，它由相邻两项之差，由某项开始奇数项取正偶数项取负求和而得，我们可以预处理出相邻的差，构造两个序列，一个奇正偶负，一个相反，然后求出最大字序列和。

#include<stdio.h>
#include<math.h>
#include<algorithm>
#define N 100100
using namespace std;

long long a[N],b[N],c[N];//数据要开到long long,不然会出错

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%lld",&a[i]);
    }
    for(int i=1; i<n; i++)
    {
        long long x=fabs(a[i]-a[i-1]);//预处理出相邻的差
        if(i%2)//一个奇正偶负的数组
        {
            b[i]=x;
        }
        else//一个偶正奇负的数组
        {
            b[i]=-x;
        }
        c[i]=-b[i];
    }
    //开始求最大子序列和
    long long s=0,maxx=0;
    for(int i=1; i<n; i++)
    {
        if(s+c[i]<0)
        {
            s=0;
        }
        else
        {
            s+=c[i];
        }
        if(s>maxx)//刷新最大值
        {
            maxx=s;
        }
    }
    s=0;
    for(int i=1; i<n; i++)
    {
        if(s+b[i]<0)
        {
            s=0;
        }
        else
        {
            s+=b[i];
        }
        if(s>maxx)//刷新最大值
        {
            maxx=s;
        }
    }
    printf("%lld\n",maxx);
    return 0;
}