#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
inline long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
inline long long power(long long a, long long n, long long m) {
  if (n == 0) return 1;
  long long p = power(a, n / 2, m);
  p = (p * p) % m;
  if (n % 2)
    return (p * a) % m;
  else
    return p;
}
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
bool dp[2001][2001];
long long dp1[2001], dp2[2001];
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  string s;
  cin >> s;
  long long n = s.length();
  for (long long i = 0; i < n; i++) dp[i][i] = 1;
  for (long long l = 2; l <= n; l++)
    for (long long i = 0; i <= n - l; i++) {
      long long j = i + l - 1;
      if (j == i + 1) {
        if (s[i] == s[j]) dp[i][j] = 1;
      } else {
        if (s[i] == s[j] && dp[i + 1][j - 1]) dp[i][j] = 1;
      }
    }
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++)
      if (dp[i][j]) dp1[i]++;
  }
  for (long long j = 0; j < n; j++)
    for (long long i = j; i >= 0; i--)
      if (dp[i][j]) dp2[j]++;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = i + 1; j < n; j++) {
      ans = ans + dp2[i] * dp1[j];
    }
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please create a solution in CPP 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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
inline long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
inline long long power(long long a, long long n, long long m) {
  if (n == 0) return 1;
  long long p = power(a, n / 2, m);
  p = (p * p) % m;
  if (n % 2)
    return (p * a) % m;
  else
    return p;
}
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
bool dp[2001][2001];
long long dp1[2001], dp2[2001];
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  string s;
  cin >> s;
  long long n = s.length();
  for (long long i = 0; i < n; i++) dp[i][i] = 1;
  for (long long l = 2; l <= n; l++)
    for (long long i = 0; i <= n - l; i++) {
      long long j = i + l - 1;
      if (j == i + 1) {
        if (s[i] == s[j]) dp[i][j] = 1;
      } else {
        if (s[i] == s[j] && dp[i + 1][j - 1]) dp[i][j] = 1;
      }
    }
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++)
      if (dp[i][j]) dp1[i]++;
  }
  for (long long j = 0; j < n; j++)
    for (long long i = j; i >= 0; i--)
      if (dp[i][j]) dp2[j]++;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = i + 1; j < n; j++) {
      ans = ans + dp2[i] * dp1[j];
    }
  }
  cout << ans << "\n";
  return 0;
}

```