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 = 2e3;
char s[MaxN + 5];
int dpl[MaxN + 5], dpr[MaxN + 5];
bool check(int x, int y) {
  while (x < y) {
    if (s[x] != s[y]) return 0;
    x++, y--;
  }
  return 1;
}
int main() {
  while (scanf("%s", s + 1) != EOF) {
    int len = strlen(s + 1);
    for (int i = 1; i <= len; i++) {
      for (int j = 1; j <= len; j++) {
        if (check(j, i + j - 1)) {
          dpl[j]++;
          dpr[i + j - 1]++;
        }
      }
    }
    long long ans = 0, sum = 0;
    for (int i = 1; i <= len; i++) {
      for (int j = 1; j <= len; j++) {
        if (j > i) {
          ans = dpr[i] * dpl[j];
          sum += ans;
        }
      }
    }
    printf("%I64d\n", sum);
    memset(dpl, 0, sizeof(dpl));
    memset(dpr, 0, sizeof(dpr));
  }
  return 0;
}