The bear has a string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (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*<sub class="lower-index">*i*</sub>*s*<sub class="lower-index">*i* + 1</sub>... *s*<sub class="lower-index">*j*</sub> 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*<sub class="lower-index">*k*</sub> = *b*, *s*<sub class="lower-index">*k* + 1</sub> = *e*, *s*<sub class="lower-index">*k* + 2</sub> = *a*, *s*<sub class="lower-index">*k* + 3</sub> = *r*.

Help the bear cope with the given problem.
The first line contains a non-empty string *s* (1 ≤ |*s*| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
import sys

def mi(s):
    return map(int, s.strip().split())

def lmi(s):
    return list(mi(s))

def mf(f, s):
    return map(f, s)

def lmf(f, s):
    return list(mf(f, s))

def main(s):
    intervals = []
    for i in range(len(s) - 3):
        if s[i:i+4] == "bear":
            intervals.append((i, i + 3))

    count = 0
    for e, inter in enumerate(intervals):
        i, j = inter
        if e == len(intervals) - 1:
            right = len(s) - j
            left = i + 1
            count += left * right
        else:
            n_i, n_j = intervals[e + 1]
            left = i + 1
            right = n_j - j 
            count += left * right
    return count

if __name__ == "__main__":
    for e, line in enumerate(sys.stdin.readlines()):
        s = line.strip()

    # Proof by induction that sorting is that
    # same as solving this problem.
    print(main(s))
