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 = 2010;
int dp[N][N];
long long pre[N];
string s;
long long power_mod(long long x, long long y) {
  x %= 1000000007;
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = (res * x) % 1000000007;
    }
    y /= 2;
    x = (x * x) % 1000000007;
  }
  return res;
}
void solve() {
  cin >> s;
  int n = s.length();
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    dp[i][i] = 1;
  }
  for (int j = 2; j <= n; j++) {
    for (int i = 1; i < j; i++) {
      if (i == (j - 1) and s[j - 1] == s[i - 1]) {
        dp[i][j] = 1;
      } else if (dp[i + 1][j - 1] == 1 and s[i - 1] == s[j - 1]) {
        dp[i][j] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    pre[i] += pre[i - 1];
    for (int j = i; j <= n; j++) {
      if (dp[i][j]) {
        ans += pre[i - 1];
        pre[j]++;
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}