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>

#define ull unsigned long long

const int Maxv = 1000010; 
const ull Prime1 = 259; 
const ull Prime2 = 213668850274628011; 
                            
int idx[Maxv], n; 
long long ans; 
ull Hash[Maxv], val[Maxv]; 
char s[Maxv]; 

std::map<ull, int> Map; 

inline void PreHandle() {
    val[n] = 1; 
    for (int i = n + 1; i <= (n << 1); i++)
        val[i] = val[i - 1] * Prime1; 
    for (int i = n - 1; i >= 0; i--)
        val[i] = val[i + 1] * Prime2; 
}

int main() {
    scanf("%d", &n); 
    scanf("%s", s); 

    PreHandle(); 

    for (int i = 0; i < n; i++) {
        Hash[i + 1] = Hash[i]; 
        idx[i + 1] = idx[i]; 

        if (s[i] == '>')
            idx[i + 1]++; 
        if (s[i] == '<')
            idx[i + 1]--;
        if (s[i] == '+')
            Hash[i + 1] += val[n + idx[i + 1]]; 
        if (s[i] == '-')
            Hash[i + 1] -= val[n + idx[i + 1]]; 
    }
    
    for (int i = n; i >= 0; i--) {
        ull tmp = Hash[i] + Hash[n] * val[n + idx[i]];
        ans += Map[tmp]; 
        Map[ Hash[i] ]++;
    }

    printf("%lld\n", ans);

    return 0; 
}