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.
s = input()
n = len(s)

# Initialize a 2D array dp to store the number of occurrences of "bear" up to index i
dp = [[0] * n for _ in range(n)]

# Initialize a variable to count the total number of occurrences of "bear"
total_occurrences = 0

# Iterate through the string s from right to left
for i in range(n - 1, -1, -1):
    
    # Update the dp array
    for j in range(i,n):
        if s[i:j+1] == 'bear':
           dp[i][j] = 1
           #print(i,j)
        elif i!=n-1 and j!= 0:
            dp[i][j] = dp[i+1][j] + dp[i][j-1]
        elif j == 0 and i != n-1:
            dp[i][j] = dp[i+1][j]
        elif i == n-1 and j != 0:
            dp[i][j] = dp[i][j-1]
        else:
            dp[i][j] = 0
for i in range(n):
    for j in range(n):
        if dp[i][j] != 0:
            total_occurrences += 1
            #print(i,j)
        
# The answer is the value in dp[3][0]
#print(dp)
print(total_occurrences)
