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.
def checkSubSequence(startIdx, endIdx, stringOne, stringTwo):
    idxOne = startIdx
    idxTwo = 0
    while(idxOne <= endIdx and idxTwo < len(stringTwo) and stringOne[idxOne] == stringTwo[idxTwo]):
        idxOne += 1
        idxTwo += 1

    return idxTwo == len(stringTwo)


def isSubString(stringOne, stringTwo, endIdx, startIdx):
    for i in range(startIdx, endIdx):
        if checkSubSequence(i, endIdx, stringOne, stringTwo):
            return True
    return False


def getCount(startIdx, endIdx, string):
    count = 0
    while(endIdx < len(string)):
        if isSubString(string, "bear", endIdx, startIdx):
            break
        endIdx += 1
    count += len(string) - endIdx
    return count


string = input()
count = 0
for startIdx in range(len(string)):
    endIdx = startIdx + 3
    if endIdx >= len(string):
        break
    count += getCount(startIdx, endIdx, string)

print(count)
