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 = 4400;
void manacher(char *s, int *d1, int *d2) {
  int len = strlen(s);
  int l = 0, r = -1;
  for (int i = 0; i < len; i++) {
    d1[i] = (r < i ? 0 : min(d1[l + r - i], r - i + 1));
    for (; i - d1[i] >= 0 && i + d1[i] < len && s[i + d1[i]] == s[i - d1[i]];
         d1[i]++)
      ;
    if (i + d1[i] - 1 > r) l = i - (d1[i] - 1), r = i + (d1[i] - 1);
  }
  l = 0, r = -1;
  for (int i = 0; i < len; i++) {
    d2[i] = (r < i ? 0 : min(d2[l + r + 1 - i], r - i + 1));
    for (; i - d2[i] - 1 >= 0 && i + d2[i] < len &&
           s[i - d2[i] - 1] == s[i + d2[i]];
         d2[i]++)
      ;
    if (i + d2[i] - 1 > r) l = i - d2[i], r = i + d2[i] - 1;
  }
}
char s[N];
int d1[N], d2[N];
long long dp[N], L[N], R[N];
int main() {
  scanf("%s", s);
  int len = strlen(s);
  manacher(s, d1, d2);
  for (int i = 0; i < len; i++) {
    int l = i - d1[i] + 1, o = i, r = i + d1[i] - 1;
    if (l) L[l - 1]--;
    L[o]++;
    R[o]++;
    R[r + 1]--;
    l = i - d2[i], o = i, r = i + d2[i] - 1;
    if (l) L[l - 1]--;
    L[o - 1]++;
    R[o]++;
    R[r + 1]--;
  }
  long long ans = 0, so_far = 0;
  for (int i = len - 1; i >= 0; i--) {
    L[i] += L[i + 1];
    dp[i] = dp[i + 1] + L[i];
  }
  for (int i = 0; i < len - 1; i++) {
    so_far += R[i];
    ans += so_far * dp[i + 1];
  }
  printf("%lld\n", ans);
  return 0;
}