Codeforces Wilbur and Array

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 positioni, 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

51 2 3 4 5

Output

5

Input

41 2 2 1

Output

3

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.

题意：给你一个数组b[]（这个数组输入值）每次可以修改数组a[]中a[i]到a[n]的值（这些数同时加1或者减1  a数组初始化为0）问你最少操作多少次可以使a数组完全等于b数组

题解：因为我们每次只能确定一个数的值（即a[i] i从1到n依次取值）所以我们应当记录下每次确定一个数需要的步数，总步数就是确定所有数的步数之和   将b数组循环遍历一遍，用sum记下操作步数，用s记录当前操作过后a[i+1]的值

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define LL long long
#define MAX 200100
using namespace std;
int b[MAX];
int main()
{
	int n,i;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<n;i++)
		    scanf("%d",&b[i]);
		__int64 s=0;
		__int64 sum=0;
		for(i=0;i<n;i++)
		{
			if(s<b[i])
			{
				sum+=(b[i]-s);//s小于b[i]则这个数需要操作b[i]-s步
				s+=(b[i]-s);//因为上边的操作为a[i]+1   +1操作所以s取值变为 +=
			}
			if(s>b[i])
			{
				sum+=(s-b[i]);//s大于b[i]则这个数需要操作s-b[i]步
				s-=(s-b[i]);//因为上边的操作为a[i]-1   -1操作所以s取值变为 -=
			}
		}
		printf("%I64d\n",sum);
	}
	return 0;
}