#include <bits/stdc++.h>
using namespace std;
char s[2001];
bool b[2000][2000];
long long sum[2000][2000];
bool p(int i, int j) {
  for (int k = i; k <= j; k++)
    if (s[k] != s[i + j - k]) return 0;
  return 1;
}
int main() {
  long long STO = 0;
  cin >> s;
  int len = strlen(s);
  for (int i = 0; i < len; i++)
    for (int j = 0; j < len; j++)
      if (i + j < len) b[j][i + j] = p(j, i + j);
  for (int i = 0; i < len; i++)
    for (int j = 0; j < len; j++)
      if (!i)
        sum[j][j] = 1;
      else if (i + j < len)
        sum[j][i + j] = sum[j][i + j - 1] + sum[j + 1][i + j] -
                        (i >= 2 ? sum[j + 1][i + j - 1] : 0) + b[j][i + j];
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++)
      STO += b[i][j] * ((i >= 1 ? sum[0][i - 1] : 0) +
                        (j + 2 <= len ? sum[j + 1][len - 1] : 0));
  cout << (STO >> 1) << endl;
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
char s[2001];
bool b[2000][2000];
long long sum[2000][2000];
bool p(int i, int j) {
  for (int k = i; k <= j; k++)
    if (s[k] != s[i + j - k]) return 0;
  return 1;
}
int main() {
  long long STO = 0;
  cin >> s;
  int len = strlen(s);
  for (int i = 0; i < len; i++)
    for (int j = 0; j < len; j++)
      if (i + j < len) b[j][i + j] = p(j, i + j);
  for (int i = 0; i < len; i++)
    for (int j = 0; j < len; j++)
      if (!i)
        sum[j][j] = 1;
      else if (i + j < len)
        sum[j][i + j] = sum[j][i + j - 1] + sum[j + 1][i + j] -
                        (i >= 2 ? sum[j + 1][i + j - 1] : 0) + b[j][i + j];
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++)
      STO += b[i][j] * ((i >= 1 ? sum[0][i - 1] : 0) +
                        (j + 2 <= len ? sum[j + 1][len - 1] : 0));
  cout << (STO >> 1) << endl;
  return 0;
}

```