#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int maxn = (int)1e5 + 10;
const int mod = (int)1e9;
int fastMax(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ y; }
int fastMin(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ x; }
vector<int> v[2005];
int main() {
  long long t, second[2005];
  long l, i, n, j;
  string a;
  cin >> a;
  n = a.size();
  second[n] = 0;
  for (i = n - 1; i >= 0; i--) {
    v[i].push_back(i);
    if (v[i + 1].size() > 0) {
      if (a[i] == a[i + 1]) v[i].push_back(i + 1);
      for (j = 0; j < v[i + 1].size(); j++)
        if (a[i] == a[v[i + 1][j] + 1]) v[i].push_back(v[i + 1][j] + 1);
    }
    second[i] = v[i].size() + second[i + 1];
  }
  t = 0;
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < v[i].size(); j++) t = t + second[v[i][j] + 1];
  }
  cout << t << endl;
  return 0;
}

### Prompt

Please formulate 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;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int maxn = (int)1e5 + 10;
const int mod = (int)1e9;
int fastMax(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ y; }
int fastMin(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ x; }
vector<int> v[2005];
int main() {
  long long t, second[2005];
  long l, i, n, j;
  string a;
  cin >> a;
  n = a.size();
  second[n] = 0;
  for (i = n - 1; i >= 0; i--) {
    v[i].push_back(i);
    if (v[i + 1].size() > 0) {
      if (a[i] == a[i + 1]) v[i].push_back(i + 1);
      for (j = 0; j < v[i + 1].size(); j++)
        if (a[i] == a[v[i + 1][j] + 1]) v[i].push_back(v[i + 1][j] + 1);
    }
    second[i] = v[i].size() + second[i + 1];
  }
  t = 0;
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < v[i].size(); j++) t = t + second[v[i][j] + 1];
  }
  cout << t << endl;
  return 0;
}

```