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).
s=raw_input()
n=len(s)
a=[]
for i in range(n-3):
    if s[i]=='b':
        if s[i:i+4]=="bear":
            a.append(i)
#print len(s)
count=0
prev=-1
for i in range(len(a)):
    if i+1<len(a):
        #print (a[i]-prev),(n-a[i]-3),(a[i+1]-a[i])
        count+=(a[i]-prev)*(n-a[i]-3)#(a[i+1]-a[i])
    else:
        #print (a[i]-prev),(n-a[i]-3)
        count+=(a[i]-prev)*(n-a[i]-3)
    #if i+1<len(a):
    #    count-=(prev+1)*(n-a[i+1]-3)
    #print (a[i]-prev+1),(n-a[i]-3)
    prev=a[i]
print count
##count=0
##for i in range(len(s)):
##    for j in range(i+1,len(s)+1):
##        if s[i:j].find("bear")>=0:
##            count+=1
##print count


##n,c=map(int,raw_input().split())
##a=map(int,raw_input().split())
##maxi=0
##for i in range(n-1):
##    diff=a[i]-a[i+1]
##    prof=diff-c
##    if prof>maxi:
##        maxi=prof
##print maxi
