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 maxn = 2000;
long long sum[maxn] = {0};
bool dp[maxn][maxn];
string s;
long long ans = 0;
int main() {
  cin >> s;
  memset(dp, false, sizeof dp);
  for (int i = 0; i < s.size(); i++) {
    dp[i][i] = true;
    sum[i]++;
    if (i + 1 >= s.size()) continue;
    if (s[i] == s[i + 1]) {
      dp[i][i + 1] = true;
      sum[i]++;
    }
  }
  for (int i = s.size() - 1; i >= 0; i--)
    for (int j = i + 2; j < s.size(); j++) {
      if (s[i] == s[j] && dp[i + 1][j - 1]) {
        dp[i][j] = true;
        sum[i]++;
      }
    }
  for (int i = s.size() - 2; i >= 0; i--) sum[i] += sum[i + 1];
  for (int i = 0; i < s.size(); i++)
    for (int j = i; j < s.size(); j++)
      if (j + 1 < s.size() && dp[i][j]) ans += sum[j + 1];
  cout << ans << endl;
  return 0;
}