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>
const int MAX_L = 2005;
const int INF_INT = 0x3f3f3f3f;
const long long INF_LL = 0x7fffffff;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const double pi = acos(-1);
using namespace std;
const int N = 5000;
char s[5000];
char t[5000];
int p[5000];
int a[5000];
int b[5000];
void manacher() {
  t[0] = '@';
  t[1] = '#';
  int n = 2;
  int tt = strlen(s + 1);
  for (int i = 1; i <= tt; i++) t[n++] = s[i], t[n++] = '#';
  s[n] = '\0';
  int mx = 0, id = 0;
  for (int i = 1; i <= n - 1; i++) {
    if (mx > i)
      p[i] = min(mx - i, p[2 * id - i]);
    else
      p[i] = 1;
    while (t[i - p[i]] == t[i + p[i]]) p[i]++;
    if (mx < i + p[i]) mx = i + p[i], id = i;
  }
}
int main(int argc, char const *argv[]) {
  while (scanf("%s", s + 1) == 1) {
    manacher();
    int len = strlen(t);
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    for (int i = 1; i < len; i++) {
      for (int j = 0; j < p[i]; j++) a[i + j]++;
    }
    for (int i = 1; i < len; i++) {
      for (int j = 0; j < p[i]; j++) b[i - j]++;
    }
    vector<int> pre, suf;
    for (int i = 2; i < len; i += 2) pre.push_back(a[i]), suf.push_back(b[i]);
    for (int i = suf.size() - 2; i >= 0; i--) suf[i] += suf[i + 1];
    long long ans = 0;
    len = strlen(s + 1);
    suf.push_back(0);
    for (int i = 1; i <= len; i++) ans += pre[i - 1] * suf[i];
    cout << ans << endl;
  }
  return 0;
}