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 re

s = input()
indexes = []
l = []
ans = 0
a = [i for i in re.split('bear', s)]
for i in range(len(s) - 3):
    if s[i:i + 4] == 'bear':
        indexes.append((i, i + 4))
for x in indexes:
    # print(s[:x[0]], s[x[1]:])
    # print(len(s[:x[0]]), len(s[x[1]:]))
    if len(s[:x[0]]) != 0 and len(s[x[1]:]) != 0:
        ans += max(len(s[:x[0]]), len(s[x[1]:])) * min(len(s[:x[0]]), len(s[x[1]:]) + 1)
    else:
        if len(s[:x[0]]) != 0:
            ans += len(s[:x[0]])
        else:
            ans += len(s[x[1]:])
    # print(s[:x[0]], s[x[1]:])
ans += 1
print(ans)
