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;
long long p[4500], dp1[4500], dp2[4500];
char b[4500];
void Manacher(char a[]) {
  memset(dp1, 0, sizeof(dp1));
  memset(dp2, 0, sizeof(dp2));
  memset(p, 0, sizeof(p));
  long long id = 0, mx = 0;
  int len = strlen(a);
  for (int i = 1; i < len; i++) {
    if (mx > i) {
      p[i] = min(p[id * 2 - i], mx - i);
    } else {
      p[i] = 1;
    }
    while (a[i - p[i]] == a[i + p[i]]) {
      p[i]++;
    }
    if (p[i] + i > mx) {
      mx = p[i] + i;
      id = i;
    }
  }
  for (int i = 1; i < len; i++) {
    for (int j = i; j < i + p[i]; j++) {
      dp1[j]++;
    }
  }
  for (int i = 1; i < len; i++) {
    for (int j = i; j > max((long long)0, i - p[i]); j--) {
      dp2[j]++;
    }
  }
  long long ans = 0;
  for (int i = 2; i < len; i += 2) {
    for (int j = i + 2; j < len; j += 2) {
      ans += dp1[i] * dp2[j];
    }
  }
  cout << ans << endl;
}
int main() {
  char a[4500];
  long long ans = 0;
  while (scanf("%s", a) != EOF) {
    b[0] = '$';
    int cnt = 1;
    int len = strlen(a);
    for (int i = 0; i < len; i++) {
      b[cnt++] = '#';
      b[cnt++] = a[i];
    }
    b[cnt++] = '#';
    b[cnt] = '\0';
    Manacher(b);
  }
  return 0;
}