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.

Examples

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).
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))
