【codeforce】C. Functions again 最大子段和

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==数学and思维==

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本文探讨了一个关于寻找特定定义下Uzhlyandian函数的最大值的问题。通过分析给定数组，利用动态规划思想，分别计算两种情况下数组元素差绝对值的最大子段和，从而得出最大值。

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:

$\text{[math]}$

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, ..., a_n (-10⁹ ≤ a_i ≤ 10⁹) — the array elements.

Output

Print the only integer — the maximum value of f.

Examples

Input

5
1 4 2 3 1

Output

Input

4
1 5 4 7

Output

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.

用两个数组存储abs(a[i]-a[i+1]),系数可正可负，取决于首位，所以用个数组分别存储，然后两个数组分别求最大子段和

注意-1的指数是i-l不是i-1

code:

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int num1[100001],num2[100001],a[100001];
int main()
{
	int n;
	while(~scanf("%d",&n)){
		for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
		for(int i=1;i<n;i++){
			if(i%2){
				num1[i]=abs(a[i]-a[i+1]);
			num2[i]=-abs(a[i]-a[i+1]);
			}
			else{
			num1[i]=-abs(a[i]-a[i+1]);
			num2[i]=abs(a[i]-a[i+1]);	
			}
		}
		long long max1=0,max2=0,sum1=0,sum2=0;
		for(int i=1;i<n;i++){
			if(sum1<0) sum1=num1[i];
			else sum1+=num1[i];
			max1=max(max1,sum1);
		}
		for(int i=1;i<n;i++){
			if(sum2<0) sum2=num2[i];
			else sum2+=num2[i];
			max2=max(max2,sum2);
		}
		printf("%I64d\n",max(max1,max2));
	}
	return 0;
 }