B - Number of Ways CodeForces - 466C

You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.

More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that .

Input

The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], ..., a[n] (|a[i]| ≤  109) — the elements of array a.

Output

Print a single integer — the number of ways to split the array into three parts with the same sum.

Examples

Input

5
1 2 3 0 3

Output

2

Input

4
0 1 -1 0

Output

1

Input

2
4 1

Output

0

题意是把给得数列分成和相等的三部分，问有几种分发。

一开始看到题就蒙了，没思路然后看了嘎哥的代码，感觉没那么难，找好n/3和n/3*2的点，前面每个N/3的点都会在遇到n/3*2的时候组成一次分发。

#include"iostream"
#include"algorithm"
#include"cstring"
using namespace std;
long long a[5*100005];
int main()
{
	int n;
	while(cin >> n)
	{
		long long x;
		cin>>a[0];
		for(int i = 1;i < n;i ++)
		{
			cin >> x;
			a[i]=a[i-1]+x;
		}
		if(a[n-1] % 3)
		{
			cout<<0<<endl;
			continue;
		}
		int res=0;
		long long  sum = 0;
		for(int i = 0;i < n-1;i ++)
		{
			if(a[i]==a[n-1]/3*2)
			{
				sum+=res;
			}
			if(a[i]==a[n-1]/3)
			{
				res++;
			} 
			
		}
		cout<<sum<<endl;
	}
	return 0;
}