#include <bits/stdc++.h>
using namespace std;
long long dp[555555], a[555555], k, l, m, n, o, p;
bool pal[5555][5555];
map<long long, long long> mp;
vector<long long> adj[555555];
const long long mod = 1e9 + 7;
string s;
int main() {
  cin >> s;
  n = s.size();
  s = ' ' + s;
  for (long long i = 1; i <= n; i++) {
    long long a = i, b = i;
    while (a >= 1 && b <= n && s[a] == s[b]) {
      pal[a][b] = 1;
      a--, b++;
    }
    a = i, b = i + 1;
    while (a >= 1 && b <= n && s[a] == s[b]) {
      pal[a][b] = 1;
      a--, b++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j++) {
      dp[i] += pal[i][j];
    }
  }
  for (long long i = n - 1; i >= 1; i--) dp[i] += dp[i + 1];
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j++) {
      if (pal[i][j]) k += dp[j + 1];
    }
  }
  cout << k;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 dp[555555], a[555555], k, l, m, n, o, p;
bool pal[5555][5555];
map<long long, long long> mp;
vector<long long> adj[555555];
const long long mod = 1e9 + 7;
string s;
int main() {
  cin >> s;
  n = s.size();
  s = ' ' + s;
  for (long long i = 1; i <= n; i++) {
    long long a = i, b = i;
    while (a >= 1 && b <= n && s[a] == s[b]) {
      pal[a][b] = 1;
      a--, b++;
    }
    a = i, b = i + 1;
    while (a >= 1 && b <= n && s[a] == s[b]) {
      pal[a][b] = 1;
      a--, b++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j++) {
      dp[i] += pal[i][j];
    }
  }
  for (long long i = n - 1; i >= 1; i--) dp[i] += dp[i + 1];
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j++) {
      if (pal[i][j]) k += dp[j + 1];
    }
  }
  cout << k;
}

```