#include <bits/stdc++.h>
using namespace std;
string s;
int n;
bool g[2005][2005];
long long f[2005], res;
void precal() {
  for (int i = 0; i < n; i++) {
    g[i][i] = true;
    if (i < n - 1 && s[i] == s[i + 1]) g[i][i + 1] = true;
  }
  for (int l = 3; l <= n; l++)
    for (int i = 0; i < n - l + 1; i++) {
      int j = i + l - 1;
      g[i][j] = (s[i] == s[j] && g[i + 1][j - 1]);
    }
  for (int i = 0; i < n; i++)
    if (g[i][n - 1]) f[i]++;
}
int main() {
  if (0) {
    freopen("a.inp", "r", stdin);
  };
  getline(cin, s);
  n = s.length();
  precal();
  for (int i = n - 2; i >= 0; i--) {
    for (int j = i; j >= 0; j--) {
      if (!g[j][i]) continue;
      f[j]++;
      res += f[i + 1];
    }
    f[i] += f[i + 1];
  }
  printf("%I64d\n", res);
}

### Prompt

Create a solution in cpp for the following problem:

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>
using namespace std;
string s;
int n;
bool g[2005][2005];
long long f[2005], res;
void precal() {
  for (int i = 0; i < n; i++) {
    g[i][i] = true;
    if (i < n - 1 && s[i] == s[i + 1]) g[i][i + 1] = true;
  }
  for (int l = 3; l <= n; l++)
    for (int i = 0; i < n - l + 1; i++) {
      int j = i + l - 1;
      g[i][j] = (s[i] == s[j] && g[i + 1][j - 1]);
    }
  for (int i = 0; i < n; i++)
    if (g[i][n - 1]) f[i]++;
}
int main() {
  if (0) {
    freopen("a.inp", "r", stdin);
  };
  getline(cin, s);
  n = s.length();
  precal();
  for (int i = n - 2; i >= 0; i--) {
    for (int j = i; j >= 0; j--) {
      if (!g[j][i]) continue;
      f[j]++;
      res += f[i + 1];
    }
    f[i] += f[i + 1];
  }
  printf("%I64d\n", res);
}

```