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 = 2005;
int n;
string s;
vector<long long> one(N, 0);
vector<long long> two(N, 0);
long long mod;
long long p;
void ini() {
  for (int i = 1; i <= n; i++) one[i] = (one[i - 1] * p + (s[i] - 'a')) % mod;
  for (int i = n; i >= 1; i--) two[i] = (two[i + 1] * p + (s[i] - 'a')) % mod;
}
long long fpow(long long b, long long e) {
  long long res = 1;
  while (e) {
    if (e & 1) res = (res * b) % mod;
    b = (b * b) % mod;
    e /= 2;
  }
  return res;
}
long long qone(int l, int r) {
  long long res = one[r];
  res = (res - one[l - 1] * fpow(p, r - l + 1)) % mod;
  res = (res + mod) % mod;
  return res;
}
long long qtwo(int l, int r) {
  long long res = two[l];
  res = (res - two[r + 1] * fpow(p, r - l + 1)) % mod;
  res = (res + mod) % mod;
  return res;
}
bool pal(int l, int r) { return (qone(l, r) == qtwo(l, r)); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  mt19937 gen(chrono::steady_clock::now().time_since_epoch().count());
  cin >> s;
  n = s.size();
  s = '?' + s;
  mod = 1000000007;
  p = 31;
  ini();
  long long res = 0;
  vector<long long> dp(n + 2, 0);
  for (int i = n; i >= 1; i--) {
    dp[i] = dp[i + 1];
    for (int j = i; j <= n; j++) {
      if (pal(i, j)) {
        res += dp[j + 1];
        dp[i]++;
      }
    }
  }
  cout << res << "\n";
  return 0;
}