#include <bits/stdc++.h>
template <typename T>
T inline SQR(const T &a) {
  return a * a;
}
template <typename T>
T inline ABS(const T &a) {
  return a < 0 ? -a : a;
}
const double EPS = 1e-9;
inline int SGN(double a) {
  return ((a > EPS) ? (1) : ((a < -EPS) ? (-1) : (0)));
}
inline int CMP(double a, double b) { return SGN(a - b); }
using namespace std;
long long bit[2222 + 1];
void up(int idx) {
  idx++;
  while (idx <= 2222) {
    bit[idx]++;
    idx += (idx & -idx);
  }
}
long long get(int idx) {
  idx++;
  long long ans = 0;
  while (idx > 0) {
    ans += bit[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  int n = int((s).size());
  bool palin[n][n];
  for (int i = 0; i < n; ++i)
    for (int j = i; j <= n - 1; ++j) palin[i][j] = false;
  for (int i = 0; i < n; ++i) {
    palin[i][i] = true;
    up(i);
  }
  for (int i = 0; i < n - 1; ++i) {
    palin[i][i + 1] = (s[i] == s[i + 1]);
    if (palin[i][i + 1]) up(i);
  }
  for (int len = 3; len <= n; ++len) {
    for (int i = 0; i < n; ++i) {
      int j = i + len - 1;
      if (j >= n) break;
      palin[i][j] = ((s[i] == s[j]) && palin[i + 1][j - 1]);
      if (palin[i][j]) up(i);
    }
  }
  long long all = get(n - 1);
  long long ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = i; j <= n - 1; ++j)
      if (palin[i][j]) ans += (all - get(j));
  cout << ans << "\n";
  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>
template <typename T>
T inline SQR(const T &a) {
  return a * a;
}
template <typename T>
T inline ABS(const T &a) {
  return a < 0 ? -a : a;
}
const double EPS = 1e-9;
inline int SGN(double a) {
  return ((a > EPS) ? (1) : ((a < -EPS) ? (-1) : (0)));
}
inline int CMP(double a, double b) { return SGN(a - b); }
using namespace std;
long long bit[2222 + 1];
void up(int idx) {
  idx++;
  while (idx <= 2222) {
    bit[idx]++;
    idx += (idx & -idx);
  }
}
long long get(int idx) {
  idx++;
  long long ans = 0;
  while (idx > 0) {
    ans += bit[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  int n = int((s).size());
  bool palin[n][n];
  for (int i = 0; i < n; ++i)
    for (int j = i; j <= n - 1; ++j) palin[i][j] = false;
  for (int i = 0; i < n; ++i) {
    palin[i][i] = true;
    up(i);
  }
  for (int i = 0; i < n - 1; ++i) {
    palin[i][i + 1] = (s[i] == s[i + 1]);
    if (palin[i][i + 1]) up(i);
  }
  for (int len = 3; len <= n; ++len) {
    for (int i = 0; i < n; ++i) {
      int j = i + len - 1;
      if (j >= n) break;
      palin[i][j] = ((s[i] == s[j]) && palin[i + 1][j - 1]);
      if (palin[i][j]) up(i);
    }
  }
  long long all = get(n - 1);
  long long ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = i; j <= n - 1; ++j)
      if (palin[i][j]) ans += (all - get(j));
  cout << ans << "\n";
  return 0;
}

```