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 INF = (int)1e9;
const long long INF64 = (long long)1e18;
long double eps = 1e-6;
const long double pi = 3.14159265358979323846;
bool p[2001][2001];
int main() {
  string s;
  cin >> s;
  for (int i = 0; i < s.length(); i++) p[i][i] = 1;
  for (int i = 0; i < s.length() - 1; i++) p[i][i + 1] = s[i] == s[i + 1];
  for (int l = 3; l <= s.length(); l++) {
    for (int i = 0; i < s.length() && ((i + l) <= s.length()); i++) {
      p[i][i + l - 1] = (s[i] == s[i + l - 1]) & p[i + 1][i + l - 2];
    }
  }
  long long sum[20001] = {0};
  sum[0] = 1;
  for (int i = 1; i < s.length(); i++) {
    sum[i] = sum[i - 1];
    for (int j = i; j >= 0; j--) {
      sum[i] += p[j][i];
    }
  }
  long long dp[2001];
  dp[0] = 0;
  for (int i = 1; i < s.length(); i++) {
    dp[i] = dp[i - 1];
    long long x = 0;
    for (int j = i; j > 0; j--) {
      x += sum[j - 1] * p[j][i];
    }
    dp[i] += x;
  }
  cout << dp[s.length() - 1];
}