#include <bits/stdc++.h>
using namespace std;
long long n, a[3000], b[3000], w[500], pos[500][5000], ans;
string second;
int ispal(int x, int y) {
  int t = y;
  while (x != y) {
    if (second[x] != second[y]) return 0;
    if (y - x == 1 || y - x == 2) {
      b[t]++;
      return 1;
    }
    x++, y--;
  }
}
int main() {
  cin >> second;
  for (int i = second.size() - 1; i >= 0; i--) {
    a[i] = a[i + 1] + 1;
    for (int j = 0; j < w[second[i] - 97]; j++) {
      a[i] += ispal(i, pos[second[i] - 97][j]);
    }
    pos[second[i] - 97][w[second[i] - 97]] = i;
    w[second[i] - 97]++;
  }
  for (int i = 0; i < second.size() - 1; i++) {
    b[i]++;
    ans += b[i] * a[i + 1];
  }
  cout << ans;
  return 0;
}

### Prompt

Generate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, a[3000], b[3000], w[500], pos[500][5000], ans;
string second;
int ispal(int x, int y) {
  int t = y;
  while (x != y) {
    if (second[x] != second[y]) return 0;
    if (y - x == 1 || y - x == 2) {
      b[t]++;
      return 1;
    }
    x++, y--;
  }
}
int main() {
  cin >> second;
  for (int i = second.size() - 1; i >= 0; i--) {
    a[i] = a[i + 1] + 1;
    for (int j = 0; j < w[second[i] - 97]; j++) {
      a[i] += ispal(i, pos[second[i] - 97][j]);
    }
    pos[second[i] - 97][w[second[i] - 97]] = i;
    w[second[i] - 97]++;
  }
  for (int i = 0; i < second.size() - 1; i++) {
    b[i]++;
    ans += b[i] * a[i + 1];
  }
  cout << ans;
  return 0;
}

```