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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int dp[2004][2004];
int n;
int dleft[2004];
int dright[2004];
long long cum[2004];
void fill(string a) {
  memset(dp, false, sizeof(dp));
  for (int i = 0; i < n; i++) {
    dp[i][i] = true;
  }
  for (int i = 0; i < n - 1; i++) {
    if (a[i] == a[i + 1]) {
      dp[i][i + 1] = true;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i + 2; j < n; j++) {
      dp[i][j] = dp[i + 1][j - 1] && (a[i] == a[j]);
    }
  }
}
void calcL() {
  memset(dleft, 0, sizeof(dleft));
  for (int i = 0; i < n; i++) {
    int s = 0;
    for (int j = 0; j <= i; j++) {
      s += dp[j][i];
    }
    dleft[i] = s;
  }
}
void calcR() {
  memset(dright, 0, sizeof(dright));
  for (int i = 0; i < n; i++) {
    int s = 0;
    for (int j = 0; j <= i; j++) {
      s += dp[j][i];
    }
    dright[i] = s;
  }
  memset(cum, 0, sizeof(cum));
  cum[0] = dright[0];
  for (int i = 1; i < n; i++) {
    cum[i] = cum[i - 1] + dright[i];
  }
}
int main() {
  string s;
  cin >> s;
  n = s.length();
  fill(s);
  calcL();
  reverse(s.begin(), s.end());
  fill(s);
  calcR();
  long long ans = 0;
  for (int i = 0; i < n - 1; i++) {
    ans += (cum[n - i - 1 - 1] * dleft[i]);
  }
  cout << ans << endl;
  return 0;
}