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;
void RunProgram();
int main() { RunProgram(); }
const int MAXN = 2005;
char s[MAXN];
bool p[MAXN][MAXN];
int n, cL[MAXN], cR[MAXN];
long long res;
void RunProgram() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  memset(p, false, sizeof p);
  for (int i = 1; i <= n; i++) p[i][i] = true;
  for (int i = 1; i < n; i++) p[i][i + 1] = (s[i] == s[i + 1]);
  for (int k = 2; k < n; k++)
    for (int i = 1; i <= n - k; i++) {
      int j = i + k;
      p[i][j] = (s[i] == s[j] && p[i + 1][j - 1]);
    }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      if (p[i][j]) ++cL[j], ++cR[i];
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) res += cL[i] * cR[j];
  cout << res;
}