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 = 0;
int dp[2001][2001];
long long int arr[2001][2001];
void compute_dp(int n) {
  for (int i = 0; i < n; i++) {
    int count = 0;
    for (int j = n - 1; j >= i; j--) {
      count += dp[i][j];
    }
    dp[i][i] = count;
  }
}
void calculate_count(int n) {
  for (int i = 0; i < n; i++) {
    arr[n - 1][i] = 0;
  }
  for (int i = n - 2; i >= 0; i--) {
    int count = 0;
    for (int j = n - 2; j >= i; j--) {
      arr[i][j] = arr[j + 1][j + 1] + dp[j + 1][j + 1];
    }
  }
  long long int count = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (dp[i][j] != 0) count += arr[i][j];
    }
  }
  cout << count;
}
void compute(char *ch) {
  int len = strlen(ch);
  n = len;
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i <= len; i++) {
    dp[i][i] = 1;
  }
  for (int n = 1; n <= len; n++) {
    for (int i = 0; i <= len - n - 1; i++) {
      int j = i + n;
      if (n == 1) {
        if (ch[i] == ch[j]) {
          dp[i][j] = 1;
        }
      } else if (ch[i] == ch[j]) {
        dp[i][j] = dp[i + 1][j - 1];
      }
    }
  }
  compute_dp(len);
  calculate_count(len);
  return;
}
int main() {
  char ch[2002];
  cin >> ch;
  compute(ch);
}