#include <bits/stdc++.h>
namespace {
char s[2111];
int c[2111][2111];
int a[2111][2111];
int a1[2111][2111];
}  // namespace
int main() {
  scanf("%s", s);
  int n = strlen(s);
  if (n < 2) {
    printf("0");
    return 0;
  }
  for (int i = 0; s[i]; ++i) {
    c[i][i] = 1;
    for (int l = i - 1, r = i + 1; s[r] && l >= 0 && s[r] == s[l]; ++r, --l)
      c[l][r] = 1;
    for (int l = i, r = i + 1; s[r] && l >= 0 && s[r] == s[l]; ++r, --l)
      c[l][r] = 1;
  }
  for (int l = 0, j; s[l]; ++l) {
    for (int i = 0; s[j = (i + l)]; ++i) {
      a[i][j] = a[i + 1][j] + c[i][j];
    }
  }
  for (int l = 0, j; s[l]; ++l) {
    for (int i = 0; s[j = (i + l)]; ++i) {
      a1[i][j] = a1[i][j - 1] + +c[i][j];
    }
  }
  long long sum = 0;
  --n;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      sum += a[0][i] * a1[j][n];
    }
  }
  std::cout << sum;
  return 0;
}

### 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>
namespace {
char s[2111];
int c[2111][2111];
int a[2111][2111];
int a1[2111][2111];
}  // namespace
int main() {
  scanf("%s", s);
  int n = strlen(s);
  if (n < 2) {
    printf("0");
    return 0;
  }
  for (int i = 0; s[i]; ++i) {
    c[i][i] = 1;
    for (int l = i - 1, r = i + 1; s[r] && l >= 0 && s[r] == s[l]; ++r, --l)
      c[l][r] = 1;
    for (int l = i, r = i + 1; s[r] && l >= 0 && s[r] == s[l]; ++r, --l)
      c[l][r] = 1;
  }
  for (int l = 0, j; s[l]; ++l) {
    for (int i = 0; s[j = (i + l)]; ++i) {
      a[i][j] = a[i + 1][j] + c[i][j];
    }
  }
  for (int l = 0, j; s[l]; ++l) {
    for (int i = 0; s[j = (i + l)]; ++i) {
      a1[i][j] = a1[i][j - 1] + +c[i][j];
    }
  }
  long long sum = 0;
  --n;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      sum += a[0][i] * a1[j][n];
    }
  }
  std::cout << sum;
  return 0;
}

```