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 MX 500000
using namespace std;
typedef long long lnt;
template <class T> inline void read(T &x) {
	x = 0; int c = getchar(), f = 1;
	for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
	for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
int n; char s[MX+5];
lnt pw[MX+5], h[MX+5];
map <lnt, int> c; lnt cur, ans;
lnt p = 233, q = 7204522363551799129;
int main() {
	read(n), scanf("%s", s+1), cur = 1;
	for (int i = 1; i <= n; c[h[i++]]++) {
		if (s[i] ==	'+') h[i] = h[i-1]+cur;
		if (s[i] == '-') h[i] = h[i-1]-cur;
		if (s[i] == '>') cur *= p, h[i] = h[i-1];
		if (s[i] == '<') cur *= q, h[i] = h[i-1];
	}
	cur = 1;
	for (int i = 1; i <= n; c[h[i++]]--) {
		ans += c[h[n]*cur+h[i-1]];
		if (s[i] == '>') cur *= p;
		if (s[i] == '<') cur *= q;
	}
	return printf("%lld\n", ans), 0;
}
