Codeforces Round #226 (Div. 2) B

B. Bear and Strings

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.

String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.

Help the bear cope with the given problem.

Input

The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.

Output

Print a single number — the answer to the problem.

Sample test(s)

input

bearbtear

output

6

input

bearaabearc

output

20

Note

In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).

In the second sample, the following pairs (i, j) match: (1,  4), (1,  5), (1,  6), (1,  7), (1,  8), (1,  9), (1,  10), (1,  11), (2,  10), (2,  11), (3,  10), (3,  11), (4,  10), (4,  11), (5,  10), (5,  11), (6,  10), (6,  11), (7,  10), (7,  11).

#include <iostream>
#include <string>
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;

int gao(string s){
    vector< pair<int,int> >seg ;
    vector< pair<int,int> >::iterator it ;
    seg.clear() ;
    int i ,Len = s.length() ,L ,R ,ans =   ,reL ;
    for(i =  ; i <= Len -  ; i++){
        if(s[i]=='b'&&s[i+]=='e'&&s[i+]=='a'&&s[i+]=='r')
            seg.push_back(make_pair(i+,i++)) ;
    }
    if(seg.size() == )
       return  ;
    it = seg.begin() ;
    reL = L = it->first  ;
    R = it->second ;
    ans += L * (Len - R +) ;
    for(it++; it != seg.end() ; it++){
        L = it->first  ;
        R = it->second ;
        ans += (L - reL) * (Len - R + ) ;
        reL = L ;
    }
    return ans ;
}

int main(){
    string s ;
    while(cin>>s){
        cout<<gao(s)<<endl ;
    }
    return  ;
}