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;
#define mod 233
int n;
char s[250005];
map<unsigned long long,int>a;
unsigned long long f[250005],g[250005],inv=1,xx=mod;
int main(){
    scanf("%d%s",&n,s+1);
    for(int i=1;i<=63;i++){
        inv=inv*xx;
        xx=xx*xx;
    }
    f[0]=0;g[0]=1;
    for(int i=1;i<=n;i++) {
        f[i]=f[i-1];g[i]=g[i-1];
        switch (s[i]){
            case '+':f[i]+=g[i];break;
            case '-':f[i]-=g[i];break;
            case '<':g[i]*=inv;break;
            case '>':g[i]*=mod;break;
        }
        a[f[i]]++;
    }
    long long ans=0;
    for(int i=1;i<=n;i++){
        ans+=a[f[n]*g[i-1]+f[i-1]];
        a[f[i]]--;
    }
    printf("%lld",ans);
    return 0;
}

