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<map>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<iostream>
    #define LLu unsigned long long
    using namespace std;
    char s[250010];
    LLu base=233,inv=7204522363551799129,f[250010],X=1,Y,ans=0;
    int n;
    map<LLu,int> mp;
    int main()
    {
    	scanf("%d",&n);getchar();
    	scanf("%s",s+1);
    	for(int i=1;i<=n;i++)
    	{
    		f[i]=f[i-1];
    		if(s[i]=='+') f[i]+=X;
    		if(s[i]=='-') f[i]-=X;
    		if(s[i]=='<') X*=base;
    		if(s[i]=='>') X*=inv;
    		mp[f[i]]++;
    	}
    	X=1;
    	for(int i=1;i<=n;i++)
    	{
    		ans+=mp[f[n]*X+Y];
    		if(s[i]=='+') Y+=X;
    		if(s[i]=='-') Y-=X;
    		if(s[i]=='<') X*=base;
    		if(s[i]=='>') X*=inv;
    		mp[f[i]]--;
    	}
    	printf("%llu",ans);
    }