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 Rint register int
using namespace std;
typedef long long LL;
typedef __int128 I128;
const int N = 250003;
const LL mod = 1000000000000000031ll, X = 1000000000000000003ll, inv = 821428571428571454ll;
int n, a[N];
LL ans, p[N], val[N];
char str[N];
map<LL, int> tong;
inline void upd(LL &a, LL b){a += b; if(a >= mod) a -= mod;}
inline LL add(LL a, LL b){return (a + b >= mod) ? (a + b - mod) : (a + b);}
int main(){
	scanf("%d%s", &n, str + 1);
	p[0] = 1;
	for(Rint i = 1;i <= n;i ++){
		p[i] = p[i - 1]; val[i] = val[i - 1];
		if(str[i] == '+') upd(val[i], p[i]);
		else if(str[i] == '-') upd(val[i], mod - p[i]);
		else if(str[i] == '<') p[i] = (I128) p[i] * inv % mod;
		else if(str[i] == '>') p[i] = (I128) p[i] * X % mod;
	}
	for(Rint i = n;i;i --){
		++ tong[val[i]];
		ans += tong[add((I128) p[i - 1] * val[n] % mod, val[i - 1])];
	}
	printf("%lld", ans);
}