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 N = 2005;
int dp[N];
char s[N];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n; ++i) {
    for (int j = i, k = i; j >= 0 && k < n && s[j] == s[k]; --j, ++k) ++dp[k];
    for (int j = i - 1, k = i; j >= 0 && k < n && s[j] == s[k]; --j, ++k)
      ++dp[k];
    if (i > 0) dp[i] += dp[i - 1];
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i, k = i; j >= 0 && k < n && s[j] == s[k]; --j, ++k)
      if (j) ans += (long long)dp[j - 1];
    for (int j = i - 1, k = i; j >= 0 && k < n && s[j] == s[k]; --j, ++k)
      if (j) ans += (long long)dp[j - 1];
  }
  printf("%I64d\n", ans);
  return 0;
}