In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475
#include <bits/stdc++.h>
using namespace std;
const unsigned long long BASE=233,INV=7204522363551799129;
const int MAXN=250100;
map<unsigned long long,int> Map;
char s[MAXN];
int n;
unsigned long long f[MAXN],X=1,Y,ans;
signed main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>n>>(s+1);
	for(int i=1;i<=n;i++)
	{
		if(s[i]=='+') f[i]=f[i-1]+X;
		if(s[i]=='-') f[i]=f[i-1]-X;
		if(s[i]=='<') f[i]=f[i-1],X=X*BASE;
		if(s[i]=='>') f[i]=f[i-1],X=X*INV;
		Map[f[i]]++;
	}
	X=1;
	for(int i=1;i<=n;i++)
	{
		ans=ans+Map[f[n]*X+Y];
		if(s[i]=='+') Y=Y+X;
		if(s[i]=='-') Y=Y-X;
		if(s[i]=='<') X=X*BASE;
		if(s[i]=='>') X=X*INV;
		Map[f[i]]--;
	}
	cout<<ans<<endl;
	return 0;
}