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 <iostream>
#include <map>

using u64 = unsigned long long;

const u64 B = 13131;
const int N = 500000 + 7;

int n;
char s[N];
u64 f[N];
int g[N];
std::map<u64, u64> map;
u64 cnt;

u64 pow(u64 a, u64 k) {
	if (k < 0) k += (1llu << 63) - 1;
	u64 t = 1;
	for (; k; a = a * a, k >>= 1)
		if (k & 1) t = t * a;
	return t;
}

int main() {
//	freopen("code.in", "r", stdin);
//	freopen("code.out", "w", stdout);

	scanf("%d%s", &n, s + 1);
	for (int i = 1; i <= n; ++i)
		if (s[i] == '+' || s[i] == '-')
			f[i] = f[i - 1] + (s[i] == '+' ? 1 : -1) * pow(B, g[i] = g[i - 1]);
		else
			f[i] = f[i - 1], g[i] = g[i - 1] + (s[i] == '<' ? -1 : 1);
	for (int i = n; i; --i)
		++map[f[i]], cnt += map[f[n] * pow(B, g[i - 1]) + f[i - 1]];
	printf("%lld\n", cnt);

	return 0;
}
