#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e3 + 5;
bool mark[maxn][maxn];
long long le[maxn], pa[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, sum = 0;
  string s;
  cin >> s;
  n = s.size();
  for (long long i = 0; i < n; i++)
    for (long long l = i, r = i; l >= 0 && r < n && s[l] == s[r]; l--, r++)
      mark[l][r] = 1, sum++;
  for (long long i = 0; i < n; i++)
    for (long long l = i, r = i + 1; l >= 0 && r < n && s[l] == s[r]; l--, r++)
      mark[l][r] = 1, sum++;
  for (long long i = 0; i < n; i++)
    for (long long j = i; j < n; j++)
      if (mark[i][j]) le[i]++;
  long long ans = 0;
  pa[0] = le[0];
  for (long long j = 1; j < n; j++) pa[j] = (pa[j - 1] + le[j]);
  for (long long i = 0; i < n; i++)
    for (long long j = i; j < n; j++)
      if (mark[i][j]) ans += (sum - pa[j] + 1000000007) % 1000000007;
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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 long long maxn = 2e3 + 5;
bool mark[maxn][maxn];
long long le[maxn], pa[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, sum = 0;
  string s;
  cin >> s;
  n = s.size();
  for (long long i = 0; i < n; i++)
    for (long long l = i, r = i; l >= 0 && r < n && s[l] == s[r]; l--, r++)
      mark[l][r] = 1, sum++;
  for (long long i = 0; i < n; i++)
    for (long long l = i, r = i + 1; l >= 0 && r < n && s[l] == s[r]; l--, r++)
      mark[l][r] = 1, sum++;
  for (long long i = 0; i < n; i++)
    for (long long j = i; j < n; j++)
      if (mark[i][j]) le[i]++;
  long long ans = 0;
  pa[0] = le[0];
  for (long long j = 1; j < n; j++) pa[j] = (pa[j - 1] + le[j]);
  for (long long i = 0; i < n; i++)
    for (long long j = i; j < n; j++)
      if (mark[i][j]) ans += (sum - pa[j] + 1000000007) % 1000000007;
  cout << ans;
  return 0;
}

```