#include <bits/stdc++.h>
using namespace std;
char s[2000 + 100];
long long dp[2000 + 100];
int d[2000 + 100];
int n;
bool is_legal(int l, int r) {
  bool ok = true;
  while (l < r) {
    if (s[l - 1] != s[r - 1]) {
      ok = false;
      break;
    }
    l++;
    r--;
  }
  return ok;
}
int main() {
  while (scanf("%s", s) == 1) {
    n = strlen(s);
    memset(dp, 0, sizeof(dp));
    memset(d, 0, sizeof(d));
    dp[1] = 1;
    d[1] = 1;
    for (int i = 2; i <= n; i++) {
      for (int j = 1; j <= i; j++) {
        if (is_legal(j, i)) {
          for (int k = 1; k <= j - 1; k++) dp[i] += d[k];
          d[i]++;
        }
      }
    }
    long long ans = 0;
    for (int i = 2; i <= n; i++) ans += dp[i];
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
char s[2000 + 100];
long long dp[2000 + 100];
int d[2000 + 100];
int n;
bool is_legal(int l, int r) {
  bool ok = true;
  while (l < r) {
    if (s[l - 1] != s[r - 1]) {
      ok = false;
      break;
    }
    l++;
    r--;
  }
  return ok;
}
int main() {
  while (scanf("%s", s) == 1) {
    n = strlen(s);
    memset(dp, 0, sizeof(dp));
    memset(d, 0, sizeof(d));
    dp[1] = 1;
    d[1] = 1;
    for (int i = 2; i <= n; i++) {
      for (int j = 1; j <= i; j++) {
        if (is_legal(j, i)) {
          for (int k = 1; k <= j - 1; k++) dp[i] += d[k];
          d[i]++;
        }
      }
    }
    long long ans = 0;
    for (int i = 2; i <= n; i++) ans += dp[i];
    printf("%I64d\n", ans);
  }
  return 0;
}

```