B. Bear and Strings

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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).

解题说明：题目的意思是找到一个包含"bear"的子串，要求bear连续出现即可，对子串的首尾没有要求。假设字符串总长为n，如果找到了bear，起始位置为i(从0开始计算)，那么一共有n-i-3个子串满足包含“bear”的条件。当然，如果后面还能找到连续的“bear”，那么为了和之前的区分，从第一次出现bear的e开始到第二次出现bear的b这个区间内都可以作为第二个包含bear的子串的起始位置。这样遍历下去即可，最后累加。

#include<iostream>#include<cstdio>#include<cmath>#include<cstdlib>#include <algorithm>#include<cstring>using namespace std;int main(){char s[5000];int i, j, k, n;long count = 0;char t;scanf("%s", &s);n = strlen(s);j = -1;for (i = 0; i<n - 3; i++){if (strncmp(s + i, "bear", 4) == 0){count += (i - j)*(n - i - 3);j = i;}}printf("%ld\n", count);return 0;}