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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
#pragma GCC optimize("fast-math")
#pragma warning(disable : 4996)
using namespace std;
const int MAXN = 30001;
const int inf = 1000000000;
const long long INF = 1ll * inf * inf;
const long double EPS = 1e-7;
const int mod7 = inf + 7;
int hp = 29;
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  ;
  string s;
  cin >> s;
  long long n = s.size();
  vector<vector<bool>> pal(n, vector<bool>(n));
  for (long long l = n - 1; l >= 0; --l) {
    for (long long r = l; r < n; ++r) {
      if (l == r)
        pal[l][r] = 1;
      else if (r - l == 1)
        pal[l][r] = (s[l] == s[r]);
      else {
        pal[l][r] = (pal[l + 1][r - 1] & (s[l] == s[r]));
      }
    }
  }
  vector<long long> sum(n);
  for (long long i = 0; i < n; ++i) {
    for (long long j = i; j >= 0; --j) {
      if (pal[j][i]) {
        sum[i]++;
      }
    }
    sum[i] += (i == 0 ? 0 : sum[i - 1]);
  }
  vector<long long> dp(n);
  for (long long i = 1; i < n; ++i) {
    for (long long j = i; j >= 1; --j) {
      dp[i] += sum[j - 1] * pal[j][i];
    }
    dp[i] += dp[i - 1];
  }
  cout << dp[n - 1];
}