Problem:
Score : 1200 points

Problem Statement

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.

Sample Input 1

5
+>+<-

Sample Output 1

3

If Takahashi eats all the symbols in S, A_1 = 1 and all other elements would be 0.
The pairs (i, j) that leads to the same sequence A are as follows:

(1, 5)

(2, 3)

(2, 4)

Sample Input 2

5
+>+-<

Sample Output 2

5

Note that the value of P may be different from the value when Takahashi eats all the symbols in S.

Sample Input 3

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<

Sample Output 3

475

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
const int N = 250001,M = 1e9 + 9,b = 1e0 + 7;
int cs[N],p[N],n;
map<int,int>mp;
void add(int &a,int b){
	a+=b;
	if(a >= M)a-=M;
	if(a < 0)a+=M;
}
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 main(){
	//freopen("readin.txt","r",stdin);
	scanf("%d",&n);
	p[0] = 1;
	char c;
	int 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] = p[i]*1LL*b%M;
		else if(c == '<')p[i] = p[i]*1LL*inv%M;
	}
	long long ans = 0;
	for(int i = 0;i <= n;i++){
		ans+=mp[cs[i]];
		mp[(cs[n]*1LL*p[i] + cs[i] + M)%M]++;
	}
	cout << ans << endl;
}