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 pre[2010], beh[2010];
long long dp[2010][2010];
char str[2010];
int n;
void init() {
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i <= n + 1; i++) {
    dp[i][0] = 1;
  }
}
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  init();
  memset(pre, 0, sizeof(pre));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      if (str[i] == str[i - j + 1]) {
        if (j < 2)
          dp[i][j] = 1;
        else {
          if (dp[i - 1][j - 2]) {
            dp[i][j] = 1;
          }
        }
      }
      pre[i] += dp[i][j];
    }
  }
  init();
  memset(beh, 0, sizeof(beh));
  for (int i = n; i >= 1; i--) {
    for (int j = 1; i + j - 1 <= n; j++) {
      if (str[i] == str[i + j - 1]) {
        if (j < 2) {
          dp[i][j] = 1;
        } else {
          if (dp[i + 1][j - 2]) {
            dp[i][j] = 1;
          }
        }
      }
      beh[i] += dp[i][j];
    }
    beh[i] += beh[i + 1];
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = ans + (pre[i] * beh[i + 1]);
  }
  printf("%lld\n", ans);
  return 0;
}