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 N = 4105;
const int M = 10005;
const int INF = 0x3f3f3f3f;
const double Pi = acos(-1);
const int MOD = 10007;
int n, m;
char s[N];
char t[N];
int Len[N];
void manacher(char *s) {
  int l = strlen(s);
  int pos = 0;
  t[pos++] = '#';
  for (int i = 0; i < l; i++) {
    t[pos++] = s[i];
    t[pos++] = '#';
  }
  t[pos] = '\0';
  l = strlen(t);
  int mx = 0;
  int id = 0;
  for (int i = 1; i < l; i++) {
    if (mx > i)
      Len[i] = min(mx - i, Len[2 * id - i]);
    else
      Len[i] = 1;
    while (t[i - Len[i]] == t[i + Len[i]] && i - Len[i] >= 0 && i + Len[i] <= l)
      Len[i]++;
    if (Len[i] + i > mx) mx = Len[i] + i, id = i;
  }
}
int pre[N], suf[N];
int main() {
  while (~scanf("%s", s)) {
    memset(Len, 0, sizeof Len);
    memset(pre, 0, sizeof pre);
    memset(suf, 0, sizeof suf);
    memset(t, 0, sizeof t);
    manacher(s);
    int l;
    l = strlen(t);
    for (int i = 1; i < l; i++)
      for (int j = 0; j < Len[i]; j++) {
        if (i + j < l) pre[i + j]++;
        if (i - j >= 0) suf[i - j]++;
      }
    l = strlen(s);
    for (int i = 0; i < l; i++) pre[i] = pre[i * 2 + 1];
    for (int i = 0; i < l; i++) suf[i] = suf[i * 2 + 1];
    for (int i = l - 2; i >= 0; i--) suf[i] += suf[i + 1];
    long long ans = 0;
    for (int i = 1; i < l; i++) ans += pre[i - 1] * suf[i];
    printf("%I64d\n", ans);
    memset(s, 0, sizeof s);
  }
  return 0;
}