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=250010;
typedef long long ll;
const ll mod=1000000000000000031ll,base=131,inv_base=68702290076335880ll;

int n;
ll p[N],q[N];
char s[N];
map<ll,int> cnt;

ll add(ll a,ll b) {
	return a+b>=mod?a+b-mod:a+b;
}

ll mul(ll a,ll b) {
	return (__int128)a*b%mod;
}

ll qpow(ll a,ll b) {
	ll ret=1;
	while(b) {
		if(b&1) ret=mul(ret,a);
		a=mul(a,a),b>>=1;
	}
	return ret;
}

int main() {
	cin>>n;scanf("%s",s+1);
	p[0]=1;//cerr<<qpow(base,mod-2)<<endl;
	for(int i=1;i<=n;i++) {
		p[i]=p[i-1],q[i]=q[i-1];
		if(s[i]=='+') q[i]=add(q[i],p[i]);
		else if(s[i]=='-') q[i]=add(q[i],mod-p[i]);
		else if(s[i]=='>') p[i]=mul(p[i],base);
		else p[i]=mul(p[i],inv_base);
	}
	ll ans=0;
	for(int i=n;i;i--) {
		++cnt[q[i]];
		ans+=cnt[add(mul(q[n],p[i-1]),q[i-1])];
	}
	cout<<ans;
	return 0;
}
