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;
int ft[2005];
long long sum(int x) {
  long long res = 0;
  while (x) {
    res += ft[x];
    x -= x & -x;
  }
  return res;
}
void update(int x) {
  while (x < 2003) {
    ft[x]++;
    x += x & -x;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  int p[n + 5][n + 5];
  for (int i = 0; i < n + 5; ++i)
    for (int j = 0; j < n + 5; ++j) p[i][j] = -INT_MAX;
  pair<int, int> pa[n * n];
  int idx = 0;
  for (int i = 0; i < n; ++i) p[i][i] = 1, pa[idx++] = {i + 2, i + 2};
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j + i < n; ++j) {
      if (s[j] == s[j + i]) {
        if (i == 1)
          p[j][j + i] = 2;
        else if (p[j + 1][j + i - 1] > 0)
          p[j][j + i] = 2 + p[j + 1][i + j - 1];
        if (p[j][j + i] > 0) pa[idx++] = {j + 2, j + i + 2};
      }
    }
  }
  sort(pa, pa + idx);
  memset(ft, 0, sizeof ft);
  long long cnt = 0;
  for (int i = 0; i < idx; ++i) {
    cnt += sum(pa[i].first - 1);
    update(pa[i].second);
  }
  cout << cnt;
}