#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  bool dp[2000][2000];
  for (int i = 0; i < 2000; i++) {
    for (int j = 1; j < 2000; j++) dp[i][j] = false;
    dp[i][0] = true;
  }
  cin >> s;
  long long st[2000], e[2000];
  for (int i = 0; i < s.length(); i++) st[i] = e[i] = 1;
  for (int i = 0; i < s.length() - 1; i++)
    if (s[i] == s[i + 1]) {
      st[i]++;
      e[i + 1]++;
      dp[i][1] = true;
    }
  for (int i = 2; i < s.length(); i++) {
    for (int j = 0; j < s.length() - i; j++) {
      if (dp[j + 1][i - 2] && s[j] == s[j + i]) {
        dp[j][i] = true;
        st[j]++;
        e[j + i]++;
      }
    }
  }
  long long res = 0;
  for (int i = s.length() - 2; i >= 0; i--) st[i] += st[i + 1];
  long long x = 0;
  for (int i = 0; i < s.length() - 1; i++) {
    res += e[i] * st[i + 1];
  }
  cout << res << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  string s;
  bool dp[2000][2000];
  for (int i = 0; i < 2000; i++) {
    for (int j = 1; j < 2000; j++) dp[i][j] = false;
    dp[i][0] = true;
  }
  cin >> s;
  long long st[2000], e[2000];
  for (int i = 0; i < s.length(); i++) st[i] = e[i] = 1;
  for (int i = 0; i < s.length() - 1; i++)
    if (s[i] == s[i + 1]) {
      st[i]++;
      e[i + 1]++;
      dp[i][1] = true;
    }
  for (int i = 2; i < s.length(); i++) {
    for (int j = 0; j < s.length() - i; j++) {
      if (dp[j + 1][i - 2] && s[j] == s[j + i]) {
        dp[j][i] = true;
        st[j]++;
        e[j + i]++;
      }
    }
  }
  long long res = 0;
  for (int i = s.length() - 2; i >= 0; i--) st[i] += st[i + 1];
  long long x = 0;
  for (int i = 0; i < s.length() - 1; i++) {
    res += e[i] * st[i + 1];
  }
  cout << res << endl;
  return 0;
}

```