Problem: 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).

Solution (in PYTHON3):

Solution:

def solve(s):

    arr = []
    n = len(s)
    for i in range(n):
        if i + 3 < n and s[i] == 'b' and s[i+1] == 'e' and s[i+2] == 'a' and s[i+3] == 'r':
            arr.append(i)

    ans = 0
    l = len(arr)
    for j in range(l):
        after = n - (arr[j] + 4)
        before = 0
        afterBefore = 0
        if j == 0:
            before = arr[j] - 0
            afterBefore = before * after
        if j > 0:
            before = arr[j] - arr[j-1] - 1
            afterBefore = before * after
        ans += after + before + afterBefore + 1

    print(ans)


# Main
def run():

    s = input()
    solve(s)

# end main


# Program Start
if __name__ == "__main__":
    run()