C. Three Parts of the Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array d1,d2,…,dnd1,d2,…,dn consisting of nn integer numbers.

Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.

Let the sum of elements of the first part be sum1sum1, the sum of elements of the second part be sum2sum2 and the sum of elements of the third part be sum3sum3. Among all possible ways to split the array you have to choose a way such that sum1=sum3sum1=sum3 and sum1sum1 is maximum possible.

More formally, if the first part of the array contains aa elements, the second part of the array contains bb elements and the third part contains cc elements, then:

sum1=∑1≤i≤adi,sum1=∑1≤i≤adi,sum2=∑a+1≤i≤a+bdi,sum2=∑a+1≤i≤a+bdi,sum3=∑a+b+1≤i≤a+b+cdi.sum3=∑a+b+1≤i≤a+b+cdi.

The sum of an empty array is 00.

Your task is to find a way to split the array such that sum1=sum3sum1=sum3 and sum1sum1 is maximum possible.

Input

The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of elements in the array dd.

The second line of the input contains nn integers d1,d2,…,dnd1,d2,…,dn (1≤di≤1091≤di≤109) — the elements of the array dd.

Output

Print a single integer — the maximum possible value of sum1sum1, considering that the condition sum1=sum3sum1=sum3 must be met.

Obviously, at least one valid way to split the array exists (use a=c=0a=c=0 and b=nb=n).

Examples

input

Copy

5
1 3 1 1 4

output

Copy

5

input

Copy

5
1 3 2 1 4

output

Copy

4

input

Copy

3
4 1 2

output

Copy

0

Note

In the first example there is only one possible splitting which maximizes sum1sum1: [1,3,1],[ ],[1,4][1,3,1],[ ],[1,4].

In the second example the only way to have sum1=4sum1=4 is: [1,3],[2,1],[4][1,3],[2,1],[4].

In the third example there is only one way to split the array: [ ],[4,1,2],[ ][ ],[4,1,2],[ ].

题意：

将一个数组分成三部分，分别为sum1 , sum2 , sum3。分成的部分可以是空。满足条件sum1 = sum3 并且sum1为最大

思路：

一开始想的是用俩个数组存下该数组从前往后和从后往前的和，然后两层for循环找答案

然后毫无疑问的超时了。。。。。

AC思路：

指针指向该数组的第一个和最后一个。sumL 和sumR分别表示从左到右和从右到左所加的和

如果sumL < sumR左指针向右++，如果sumL > sumR右指针向左++，如果相等则把这个值记录下来

当L > R时，xun'循环结束。

记得开 long long

下面代码：

#include <stdio.h>
#include <iostream>
#include <set>
#include <bits/stdc++.h>
const int maxn = 2e5+8;
using namespace std;
int main()
{
	long long int n, a[maxn] , suml ,sumr , max = -999999;
	cin >> n;
	for(int i = 1 ; i <= n ; i++)
	{
		cin >> a[i];
	}
	int l = 1 , r = n;
	suml = 0;
	sumr = 0;
		while(l <= r)
		{
			if(suml < sumr)
			{
				suml += a[l];
				l++;
			}
			else if( suml > sumr)
			{
				sumr += a[r];
				r--;
			}
			else
			{
				if(sumr > max)
				{
					max = sumr;
				}
				
				suml+=a[l];
				l++;
			}
		}
		if(suml == sumr)
		{
			if(sumr > max)
			{
				max = sumr;
			}
		}
	cout << max << endl;
	return 0;
}