#include <bits/stdc++.h>
using namespace std;
char s[2010];
int len;
int l[2010];
int r[2010];
bool dp[2010][2010];
long long ret;
int main() {
  scanf("%s", s);
  len = strlen(s);
  memset(dp, 0, sizeof(dp));
  memset(l, 0, sizeof(l));
  memset(r, 0, sizeof(r));
  for (int i = 0; i < len; ++i) dp[i][i] = true;
  for (int i = 0; i < len - 1; ++i)
    if (s[i] == s[i + 1]) dp[i][i + 1] = true;
  for (int k = 2; k < len; ++k) {
    for (int i = 0; i + k < len; ++i) {
      dp[i][i + k] = (s[i] == s[i + k]) && dp[i + 1][i + k - 1];
    }
  }
  for (int i = 0; i < len; ++i) {
    for (int j = i; j >= 0; --j) {
      l[i] += dp[j][i];
    }
    for (int j = i; j < len; ++j) {
      r[i] += dp[i][j];
    }
  }
  ret = 0LL;
  for (int i = 0; i < len; ++i) {
    for (int j = i + 1; j < len; ++j) {
      ret += (long long)l[i] * (long long)r[j];
    }
  }
  cout << ret << 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[2010];
int len;
int l[2010];
int r[2010];
bool dp[2010][2010];
long long ret;
int main() {
  scanf("%s", s);
  len = strlen(s);
  memset(dp, 0, sizeof(dp));
  memset(l, 0, sizeof(l));
  memset(r, 0, sizeof(r));
  for (int i = 0; i < len; ++i) dp[i][i] = true;
  for (int i = 0; i < len - 1; ++i)
    if (s[i] == s[i + 1]) dp[i][i + 1] = true;
  for (int k = 2; k < len; ++k) {
    for (int i = 0; i + k < len; ++i) {
      dp[i][i + k] = (s[i] == s[i + k]) && dp[i + 1][i + k - 1];
    }
  }
  for (int i = 0; i < len; ++i) {
    for (int j = i; j >= 0; --j) {
      l[i] += dp[j][i];
    }
    for (int j = i; j < len; ++j) {
      r[i] += dp[i][j];
    }
  }
  ret = 0LL;
  for (int i = 0; i < len; ++i) {
    for (int j = i + 1; j < len; ++j) {
      ret += (long long)l[i] * (long long)r[j];
    }
  }
  cout << ret << endl;
  return 0;
}

```