#include <bits/stdc++.h>
using namespace std;
const int N = 250001;
const long long M = (long double)1e18 + 31,b = 1e9 + 7;
long long cs[N],p[N],n;
map<long long,int>mp;
void add(long long &a,long long b){
	a+=b;
	if(a >= M)a-=M;
	if(a < 0)a+=M;
}
long long power(long long base,long long to){
	long long ret = 1;
	while(to){
		if(to&1)ret = (__int128)ret*base%M;
		to>>=1;
		base = (__int128)base*base%M;
	}
	return ret;
}
int main(){
//	freopen("readin.txt","r",stdin);
	scanf("%d",&n);
	p[0] = 1;
	char c;
	long long inv = power(b,M - 2);
	for(int i = 1;i <= n;i++){
		scanf(" %c",&c);
		cs[i] = cs[i - 1];
		p[i] = p[i - 1];
		if(c == '+')add(cs[i],p[i]);
		else if(c == '-')add(cs[i],-p[i]);
		else if(c == '>')p[i] = (__int128)p[i]*b%M;
		else if(c == '<')p[i] = (__int128)p[i]*inv%M;
	}
	long long ans = 0;
	for(int i = 0;i <= n;i++){
		ans+=mp[cs[i]];
		mp[((__int128)cs[n]*p[i]%M + cs[i] + M)%M]++;
	}
	cout << ans << endl;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 250001;
const long long M = (long double)1e18 + 31,b = 1e9 + 7;
long long cs[N],p[N],n;
map<long long,int>mp;
void add(long long &a,long long b){
	a+=b;
	if(a >= M)a-=M;
	if(a < 0)a+=M;
}
long long power(long long base,long long to){
	long long ret = 1;
	while(to){
		if(to&1)ret = (__int128)ret*base%M;
		to>>=1;
		base = (__int128)base*base%M;
	}
	return ret;
}
int main(){
//	freopen("readin.txt","r",stdin);
	scanf("%d",&n);
	p[0] = 1;
	char c;
	long long inv = power(b,M - 2);
	for(int i = 1;i <= n;i++){
		scanf(" %c",&c);
		cs[i] = cs[i - 1];
		p[i] = p[i - 1];
		if(c == '+')add(cs[i],p[i]);
		else if(c == '-')add(cs[i],-p[i]);
		else if(c == '>')p[i] = (__int128)p[i]*b%M;
		else if(c == '<')p[i] = (__int128)p[i]*inv%M;
	}
	long long ans = 0;
	for(int i = 0;i <= n;i++){
		ans+=mp[cs[i]];
		mp[((__int128)cs[n]*p[i]%M + cs[i] + M)%M]++;
	}
	cout << ans << endl;
}

```