#include <bits/stdc++.h>
int l, pal[2002][2002];
using namespace std;
long long sum[2002], dp[2002];
int main() {
  string s;
  cin >> s;
  l = (int)(s.size());
  dp[0] = 0;
  for (int i = (0); i <= (l - 1); i++) {
    int p = i, rp = 0;
    while (p - rp >= 0 && p + rp < l && s[p - rp] == s[p + rp]) {
      pal[p - rp][p + rp] = 1;
      rp++;
    }
    rp = 0;
    p = i;
    while (p - rp >= 0 && p + rp + 1 < l && s[p - rp] == s[p + rp + 1]) {
      pal[p - rp][p + rp + 1] = 1;
      rp++;
    }
  }
  sum[0] = 1;
  for (int i = (1); i <= (l - 1); i++) {
    sum[i] = sum[i - 1];
    for (int j = (0); j <= (i); j++)
      if (pal[j][i] == 1) sum[i]++;
  }
  for (int i = (1); i <= (l - 1); i++) {
    dp[i] = dp[i - 1];
    for (int j = (1); j <= (i); j++) dp[i] += pal[j][i] * sum[j - 1];
  }
  cout << dp[l - 1] << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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>
int l, pal[2002][2002];
using namespace std;
long long sum[2002], dp[2002];
int main() {
  string s;
  cin >> s;
  l = (int)(s.size());
  dp[0] = 0;
  for (int i = (0); i <= (l - 1); i++) {
    int p = i, rp = 0;
    while (p - rp >= 0 && p + rp < l && s[p - rp] == s[p + rp]) {
      pal[p - rp][p + rp] = 1;
      rp++;
    }
    rp = 0;
    p = i;
    while (p - rp >= 0 && p + rp + 1 < l && s[p - rp] == s[p + rp + 1]) {
      pal[p - rp][p + rp + 1] = 1;
      rp++;
    }
  }
  sum[0] = 1;
  for (int i = (1); i <= (l - 1); i++) {
    sum[i] = sum[i - 1];
    for (int j = (0); j <= (i); j++)
      if (pal[j][i] == 1) sum[i]++;
  }
  for (int i = (1); i <= (l - 1); i++) {
    dp[i] = dp[i - 1];
    for (int j = (1); j <= (i); j++) dp[i] += pal[j][i] * sum[j - 1];
  }
  cout << dp[l - 1] << endl;
  return 0;
}

```