Educational Codeforces Round 74 (Rated for Div. 2)D. AB-string

D. AB-string

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The string t1t2…tkt1t2…tk is good if each letter of this string belongs to at least one palindrome of length greater than 1.

A palindrome is a string that reads the same backward as forward. For example, the strings A, BAB, ABBA, BAABBBAAB are palindromes, but the strings AB, ABBBAA, BBBA are not.

Here are some examples of good strings:

• tt = AABBB (letters t1t1, t2t2 belong to palindrome t1…t2t1…t2 and letters t3t3, t4t4, t5t5 belong to palindrome t3…t5t3…t5);
• tt = ABAA (letters t1t1, t2t2, t3t3 belong to palindrome t1…t3t1…t3 and letter t4t4 belongs to palindrome t3…t4t3…t4);
• tt = AAAAA (all letters belong to palindrome t1…t5t1…t5);

You are given a string ss of length nn, consisting of only letters A and B.

You have to calculate the number of good substrings of string ss.

Input

The first line contains one integer nn (1≤n≤3⋅1051≤n≤3⋅105) — the length of the string ss.

The second line contains the string ss, consisting of letters A and B.

Output

Print one integer — the number of good substrings of string ss.

Examples

input

Copy

5
AABBB

output

Copy

6

input

Copy

3
AAA

output

Copy

3

input

Copy

7
AAABABB

output

Copy

15

Note

In the first test case there are six good substrings: s1…s2s1…s2, s1…s4s1…s4, s1…s5s1…s5, s3…s4s3…s4, s3…s5s3…s5 and s4…s5s4…s5.

In the second test case there are three good substrings: s1…s2s1…s2, s1…s3s1…s3 and s2…s3s2…s3.

随便在纸上写一写我们就可以发现其实不满足要求的只有一种字符串

那就使abbbbbbbbbbb或aaaaaaaaaaaab

在某一测出现了单独字符 

所以问题就转化成了统计以上字符串的个数

我们开一个vector记录相邻的相同项

用全部的数减去不合法数目即可

#include<bits/stdc++.h>
using namespace std;
char s[300005];
vector<int>ve;
int main()
{
    int n;
    scanf("%d",&n);
    scanf("%s",s+1);
    int pos=1;
    for(int i=2;i<=n;i++)
    {
        if(s[i]==s[i-1])
        {
            pos++;
        }
        else
        {
            ve.push_back(pos);
            pos=1;
        }
        if(i==n) ve.push_back(pos);
    }
    long long ans=1LL*n*(n-1)/2;
    if(ve.size()>=2)
    {
        for(int i=0;i<ve.size();i++)
        {
            if(i==0||i==ve.size()-1) ans-=ve[i];
            else ans-=ve[i]*2;
        }
        ans+=ve.size()-1;
    }
    printf("%I64d\n",ans);
}