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 int N = 250001,M = 1e9 + 9,M1 = 1e9 + 7,b = 103,b1 = 11;
int cs[N],cs1[N],p[N],p1[N],n;
map<pair<int,int>,int>mp;
void add(int &a,int b){
	a+=b;
	if(a >= M)a-=M;
	if(a < 0)a+=M;
}
void add1(int &a,int b){
	a+=b;
	if(a >= M1)a-=M1;
	if(a < 0)a+=M1;
}
int power(int base,int to){
	int ret = 1;
	while(to){
		if(to&1)ret = ret*1LL*base%M;
		to>>=1;
		base = base*1LL*base%M;
	}
	return ret;
}
int power1(int base,int to){
	int ret = 1;
	while(to){
		if(to&1)ret = ret*1LL*base%M1;
		to>>=1;
		base = base*1LL*base%M1;
	}
	return ret;
}
int main(){
//	freopen("readin.txt","r",stdin);
	scanf("%d",&n);
	p[0] = p1[0] = 1;
	char c;
	int inv = power(b,M - 2),inv1 = power1(b1,M1 - 2);
	for(int i = 1;i <= n;i++){
		scanf(" %c",&c);
		cs[i] = cs[i - 1];
		p[i] = p[i - 1];
		cs1[i] = cs1[i - 1];
		p1[i] = p1[i - 1];
		if(c == '+')add(cs[i],p[i]),add1(cs1[i],p1[i]);
		else if(c == '-')add(cs[i],-p[i]),add1(cs1[i],-p1[i]);
		else if(c == '>')p[i] = p[i]*1LL*b%M,p1[i] = p1[i]*1LL*b1%M1;
		else if(c == '<')p[i] = p[i]*1LL*inv%M,p1[i] = p1[i]*1LL*inv1%M1;
	}
	long long ans = 0;
	for(int i = 0;i <= n;i++){
		ans+=mp[{cs[i],cs1[i]}];
		mp[{(cs[n]*1LL*p[i] + cs[i] + M)%M,(cs1[n]*1LL*p1[i] + cs1[i] + M1)%M1}]++;
	}
	cout << ans << endl;
}
