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 mx = 2e3 + 12;
const int inf = 1e9 + 123;
long long f[mx], p[mx];
string s;
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    f[i] = 0;
    p[i] = 1;
  }
  queue<long long> que;
  que.push(0);
  for (int i = 1; i < s.size(); i++) {
    que.push(i);
    long long cs = que.size();
    for (int j = 1; j <= cs; j++) {
      long long tj = que.front();
      que.pop();
      if (tj != 0 && s[i] == s[tj - 1]) {
        p[i]++;
        que.push(tj - 1);
      }
    }
    p[i] += p[i - 1];
    que.push(i);
  }
  queue<long long> que1;
  que1.push(0);
  for (int i = 1; i < s.size(); i++) {
    que1.push(i);
    f[i] += f[i - 1] + p[i - 1];
    long long cs = que1.size();
    for (int j = 1; j <= cs; j++) {
      long long tj = que1.front();
      que1.pop();
      if (tj != 0 && s[i] == s[tj - 1]) {
        f[i] += p[tj - 2];
        que1.push(tj - 1);
      }
    }
    que1.push(i);
  }
  cout << f[s.size() - 1];
  return 0;
}