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;
const int maxN = 2000 + 10;
int dp[maxN][2];
bool pal[maxN][maxN];
int main() {
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) pal[i][i] = true;
  for (int i = 1; i < n; i++)
    if (s[i] == s[i - 1]) pal[i - 1][i] = true;
  for (int len = 3; len <= n; len++) {
    int i = 0, j = i + len - 1;
    for (j; j < n; j++, i++) pal[i][j] |= (s[i] == s[j] && pal[i + 1][j - 1]);
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= i; j++) dp[i][0] += pal[j][i];
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) dp[i][1] += pal[i][j];
  long long res = 0;
  for (int i = 1; i < n; i++)
    for (int j = 0; j < i; j++) res += dp[i][1] * dp[j][0];
  cout << res << endl;
  return 0;
}