Codeforce Round #226 Div2 B

   
     
 

            

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Codeforces Round #226 (Div. 2)

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Submission	Time	Verdict
5794000	01/24/2014 08:52PM	Accepted

 

 

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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) = s_is_i + 1... s_j 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 s_k = b, s_k + 1 = e, s_k + 2 = a, s_k + 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).

            

Codeforces (c) Copyright 2010-2014 Mike Mirzayanov

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Server time: 01/26/2014 10:37PM (p1).

 

 

 

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff 
#define mod 1000000007
#define ll long long
#define maxn 5025
#define pi acos(-1.0)
int n, m;
char s[maxn];
int a[maxn][maxn];
int main(){
    scanf("%s", s);
    n = strlen(s);
    int x = -1;
    for (int i = 0; i < n; i++){
        if (s[i] == 'b'&&s[i + 1] == 'e'&&s[i + 2] == 'a'&&s[i + 3] == 'r'){
            for (int k = x + 1; k <= i; k++){
                for (int j = i + 3; j < n; j++){
                    a[k][j] = 1;
                }
            }
            x = i;
        }
    }
    int t = 0;
    for (int i = 0; i < n;i++)
    for (int j = i+3; j < n; j++){
        if (a[i][j])t++;
    }
    printf("%d\n", t);
    return 0;
}