CodeForces ---596B--Wilbur and Array(贪心模拟)

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Description

Wilbur the pig is tinkering with arrays again. He has the array a₁, a₂, ..., a_n initially consisting of n zeros. At one step, he can choose any index i and either add 1 to all elements a_i, a_i + 1, ... , a_n or subtract 1 from all elements a_i, a_i + 1, ..., a_n. His goal is to end up with the array b₁, b₂, ..., b_n.

Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the length of the array a_i. Initially a_i = 0 for every position i, so this array is not given in the input.

The second line of the input contains n integers b₁, b₂, ..., b_n ( - 10⁹ ≤ b_i ≤ 10⁹).

Output

Print the minimum number of steps that Wilbur needs to make in order to achieve a_i = b_i for all i.

Sample Input

Input

5
1 2 3 4 5

Output

Input

4
1 2 2 1

Output

Hint

In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.

In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1.

Source

Codeforces Round #331 (Div. 2)
题意：
给你一个长度为n的数组，然后对一个数组a操作，每次选取一个i，对i--n的元素进行加一或者减一的操作，问至少要多少步才可以将a变为目标数组
直接贪心模拟，num[0]=0;第一个数num[1]需要操作的次数一定是加num[1]次，后边的数在他前边数的基础上操作| num[i]-num[i-1] | 次，加或者减

#include<stdio.h>
#include<string.h>
#include<math.h>
int num[200000+10];
int main()
{
	int n;
	long long sum=0;//数据范围略大，操作次数多了点 
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&num[i]);
		sum+=abs(num[i]-num[i-1]);
	}
	printf("%lld
",sum);
	return 0;
}