#include <bits/stdc++.h>
char s[2005];
int dp[2005][2005];
int main() {
  scanf("%s", s);
  int len = strlen(s);
  for (int i = 0; i < 2 * len - 1; ++i)
    for (int j = i % 2; (i - j) / 2 >= 0 && s[(i - j) / 2] == s[(i + j) / 2];
         j += 2)
      dp[(i - j) / 2][(i + j) / 2] = 1;
  for (int i = 0; i < len; ++i)
    for (int j = i + 1; j < len; ++j) dp[i][j] += dp[i][j - 1];
  long long ans = 0;
  for (int i = 0; i < len; ++i)
    for (int j = i + 1; j < len; ++j) {
      long long cnt;
      cnt = dp[i][j - 1] * dp[j][len - 1];
      ans += cnt;
    }
  printf("%I64d\n", ans);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.

In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.

A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.

A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".

Input

The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.

Output

Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.

Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.

Examples

Input

aa


Output

1


Input

aaa


Output

5


Input

abacaba


Output

36

### Response

```cpp
#include <bits/stdc++.h>
char s[2005];
int dp[2005][2005];
int main() {
  scanf("%s", s);
  int len = strlen(s);
  for (int i = 0; i < 2 * len - 1; ++i)
    for (int j = i % 2; (i - j) / 2 >= 0 && s[(i - j) / 2] == s[(i + j) / 2];
         j += 2)
      dp[(i - j) / 2][(i + j) / 2] = 1;
  for (int i = 0; i < len; ++i)
    for (int j = i + 1; j < len; ++j) dp[i][j] += dp[i][j - 1];
  long long ans = 0;
  for (int i = 0; i < len; ++i)
    for (int j = i + 1; j < len; ++j) {
      long long cnt;
      cnt = dp[i][j - 1] * dp[j][len - 1];
      ans += cnt;
    }
  printf("%I64d\n", ans);
}

```