There are n banks in the city where Vasya lives, they are located in a circle, such that any two banks are neighbouring if their indices differ by no more than 1. Also, bank 1 and bank n are neighbours if n > 1. No bank is a neighbour of itself.

Vasya has an account in each bank. Its balance may be negative, meaning Vasya owes some money to this bank.

There is only one type of operations available: transfer some amount of money from any bank to account in any neighbouring bank. There are no restrictions on the size of the sum being transferred or balance requirements to perform this operation.

Vasya doesn’t like to deal with large numbers, so he asks you to determine the minimum number of operations required to change the balance of each bank account to zero. It’s guaranteed, that this is possible to achieve, that is, the total balance of Vasya in all banks is equal to zero.

Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of banks.

The second line contains n integers ai ( - 109 ≤ ai ≤ 109), the i-th of them is equal to the initial balance of the account in the i-th bank. It’s guaranteed that the sum of all ai is equal to 0.

Output
Print the minimum number of operations required to change balance in each bank to zero.

Examples
Input
3
5 0 -5
Output
1
Input
4
-1 0 1 0
Output
2
Input
4
1 2 3 -6
Output
3
这道题大致意思就是相邻的银行之间可以转钱，求让所有银行存款为0的最小步数。
看了大佬的代码才知道，是道思维题，一个区间和最大的操作数是n-1,将它分为两个区间和为0的操作数就是n-2，…，k个区间和就是n-k，求k的最大值。求前缀和，用map来记录出现次数，该值如果出现两次，就说明中间有个区间为0，大佬的代码tql，膜拜膜拜。
AC代码：

#include<stdio.h>
#include<algorithm>
#include<map>
using namespace std;
map<long long,int>mp;
int main()
{
   
	mp.clear();
	long long s=0;
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
   
	scanf("%d",&a[i]);
	}	
		int k=0;
		for(int i=1;i<=n;i++)
		{
   	
			s+=a[i];
			mp[s]++;
		 k=max(mp[s],k);
		}
		
	printf("%d\n",n-k-1+1);
}