#include <bits/stdc++.h>
int main() {
  char s[5010], *p, *q, *f, *p1, *q1, *f1, *u;
  int i, k, t, sum, len, num;
  while (scanf("%s", s) != EOF) {
    sum = 0;
    k = 0;
    len = strlen(s);
    q = &s[len - 1];
    u = f = s;
    while (1) {
      num = 0;
      u = p = strstr(u, "bear");
      if (p == NULL) break;
      if (k == 1) {
        num = num + q - (p + 3) + 1;
        num = num + num * (p - q1 - 1);
        sum += num;
      } else {
        num = num + q - (p + 3) + 1;
        num = num + num * (p - f);
        sum = sum + num;
      }
      q1 = p;
      k = 1;
      u += 4;
    }
    printf("%d\n", sum);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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).

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  char s[5010], *p, *q, *f, *p1, *q1, *f1, *u;
  int i, k, t, sum, len, num;
  while (scanf("%s", s) != EOF) {
    sum = 0;
    k = 0;
    len = strlen(s);
    q = &s[len - 1];
    u = f = s;
    while (1) {
      num = 0;
      u = p = strstr(u, "bear");
      if (p == NULL) break;
      if (k == 1) {
        num = num + q - (p + 3) + 1;
        num = num + num * (p - q1 - 1);
        sum += num;
      } else {
        num = num + q - (p + 3) + 1;
        num = num + num * (p - f);
        sum = sum + num;
      }
      q1 = p;
      k = 1;
      u += 4;
    }
    printf("%d\n", sum);
  }
  return 0;
}

```