CodeForces-1238D-AB-string CodeForces-思维+字符串反向思考

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

5
AABBB

Output

6

Input

3
AAA

Output

3

Input

7
AAABABB

Output

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.

 1 #include<stdio.h>
 2 #include<string.h>
 3 const int N=300020;
 4 typedef long long ll;
 5
 6 char a[N],b[N];
 7
 8 int main()
 9 {
10     ll n;
11     while(~scanf("%lld",&n))
12     {
13         scanf("%s",a);
14         ll sum1=0,sum2=0,ss=0;
15         int k1=0,k2=0;
16         for(int i=1;i<n;i++)
17         {
18             if(a[i]!=a[i-1])
19             {
20                 ss++;
21                 sum1=sum1+i-k1;
22                 k1=i;
23             }
24         }
25
26         int p=0;
27         for(int i=n-1;i>=0;i--)
28             b[p++]=a[i];
29         for(int i=0;i<p;i++)
30         {
31             if(b[i]!=b[i-1])
32             {
33                 sum2=sum2+i-k2;
34                 k2=i;
35             }
36         }
37         ll w=n*(n-1)/2-sum1-sum2+ss;
38         printf("%lld\n",w);
39     }
40     return 0;
41 }