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 n, N = 0;
char S[11000], newS[11000];
int R[11000];
long long int f[11000], newf[11000], sum[11000], presum[11000];
void Manacher() {
  for (int i = 1; i < n; i++) {
    newS[++N] = S[i];
    newS[++N] = '#';
  }
  newS[++N] = S[n];
  newS[0] = '@', newS[N + 1] = '$';
  int mx = 1, id = 1;
  R[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (mx >= i)
      R[i] = min(R[2 * id - i], mx - i + 1);
    else
      R[i] = 1;
    while (newS[i + R[i]] == newS[i - R[i]]) R[i]++;
    if (i + R[i] - 1 > mx) {
      mx = i + R[i] - 1;
      id = i;
    }
  }
  for (int i = 1; i <= N; i++) f[i]++, f[i + R[i]]--;
  for (int i = 1; i <= N; i++) f[i] += f[i - 1];
  for (int i = 1, j = 1; i <= n; i++, j += 2) newf[i] = f[j];
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + newf[i];
  for (int i = 1; i <= n; i++) presum[i] = presum[i - 1] + sum[i];
  long long int ans = 0;
  for (int i = 1; i <= N; i++) {
    if (newS[i] == '#' && R[i] > 1) {
      int r = (i + 1) / 2 - 1;
      int l = (i - R[i] + 1) / 2;
      ans += presum[r] - presum[max(0, l - 1)];
    } else if (newS[i] != '#') {
      int r = (i + 1) / 2 - 1;
      int l = (i - R[i] + 1) / 2;
      ans += presum[r] - presum[max(0, l - 1)];
    }
  }
  printf("%I64d\n", ans);
}
int main() {
  scanf("%s", S + 1);
  n = strlen(S + 1);
  Manacher();
  return 0;
}