#include <bits/stdc++.h>
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
using namespace std;
string str;
bool palin[2010][2010], done[2010][2010];
long long ending[2010], starting[2010], cending[2010], cstarting[2010];
bool palingo(int start, int end) {
  if (done[start][end]) return palin[start][end];
  done[start][end] = true;
  if (start == end)
    palin[start][end] = true;
  else if (start + 1 == end)
    palin[start][end] = str[start] == str[end];
  else
    palin[start][end] = str[start] == str[end] && palingo(start + 1, end - 1);
  if (palin[start][end]) {
    starting[start]++;
    ending[end]++;
  }
  return palin[start][end];
}
int main(int argc, char const *argv[]) {
  cin >> str;
  int strl = str.size();
  memset(palin, 0, sizeof(palin));
  memset(done, 0, sizeof(done));
  memset(starting, 0, sizeof(starting));
  memset(ending, 0, sizeof(ending));
  for (int i = 0; i < strl; i++)
    for (int j = 0; j < strl; j++) palingo(i, j);
  cstarting[strl - 1] = starting[strl - 1];
  for (int i = strl - 2; i >= 0; i--)
    cstarting[i] = cstarting[i + 1] + starting[i];
  long long ans = 0;
  for (int i = 0; i < strl - 1; i++)
    if (ending[i] && cstarting[i + 1]) ans += ending[i] * cstarting[i + 1];
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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>
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
using namespace std;
string str;
bool palin[2010][2010], done[2010][2010];
long long ending[2010], starting[2010], cending[2010], cstarting[2010];
bool palingo(int start, int end) {
  if (done[start][end]) return palin[start][end];
  done[start][end] = true;
  if (start == end)
    palin[start][end] = true;
  else if (start + 1 == end)
    palin[start][end] = str[start] == str[end];
  else
    palin[start][end] = str[start] == str[end] && palingo(start + 1, end - 1);
  if (palin[start][end]) {
    starting[start]++;
    ending[end]++;
  }
  return palin[start][end];
}
int main(int argc, char const *argv[]) {
  cin >> str;
  int strl = str.size();
  memset(palin, 0, sizeof(palin));
  memset(done, 0, sizeof(done));
  memset(starting, 0, sizeof(starting));
  memset(ending, 0, sizeof(ending));
  for (int i = 0; i < strl; i++)
    for (int j = 0; j < strl; j++) palingo(i, j);
  cstarting[strl - 1] = starting[strl - 1];
  for (int i = strl - 2; i >= 0; i--)
    cstarting[i] = cstarting[i + 1] + starting[i];
  long long ans = 0;
  for (int i = 0; i < strl - 1; i++)
    if (ending[i] && cstarting[i + 1]) ans += ending[i] * cstarting[i + 1];
  cout << ans << endl;
  return 0;
}

```