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;
string str;
long long int LTR[2010], RTL[2010], sz;
bool pal[2010][2010];
long long int f();
void setPal();
int main() {
  cin >> str;
  sz = str.size();
  memset(pal, 0, sizeof(pal));
  setPal();
  cout << f() << endl;
  return 0;
}
long long int f() {
  long long int res = 0;
  LTR[0] = 1;
  RTL[sz - 1] = 1;
  if (sz == 1) {
    return 0;
  }
  for (int i = 1; i < sz; i++) {
    for (int j = 0; j <= i; j++) {
      if (pal[j][i]) {
        LTR[i]++;
      }
    }
  }
  for (int i = sz - 2; i > 0; i--) {
    RTL[i] = RTL[i + 1];
    for (int j = sz - 1; j >= i; j--) {
      if (pal[i][j]) {
        RTL[i]++;
      }
    }
  }
  for (int i = 0; i < sz - 1; i++) {
    res += LTR[i] * RTL[i + 1];
  }
  return res;
}
void setPal() {
  for (int i = 0; i < sz; i++) {
    pal[i][i] = true;
  }
  for (int i = 2; i <= sz; i++) {
    for (int j = 0; j + i <= sz; j++) {
      if (i == 2 && str[j] == str[j + i - 1]) {
        pal[j][j + i - 1] = true;
      } else if (i > 2 && pal[j + 1][j + i - 2] && str[j] == str[j + i - 1]) {
        pal[j][j + i - 1] = true;
      }
    }
  }
  return;
}