#include <bits/stdc++.h>
using namespace std;
const int maxn = 2000 + 5;
int p[maxn][maxn];
char ch[maxn];
int n;
long long dp[maxn];
long long sum[maxn];
int main() {
  scanf("%s", ch);
  int lo = 0;
  for (int i = 0; ch[i] != '\0'; i++) {
    lo++;
  }
  for (int i = 0; i < lo; i++) {
    p[i][i] = 1;
    for (int l = i - 1, r = i + 1; l >= 0 && r < lo; l--, r++) {
      if (ch[l] != ch[r]) break;
      p[l][r] = 1;
    }
    for (int l = i, r = i + 1; l >= 0 && r < lo; l--, r++) {
      if (ch[l] != ch[r]) break;
      p[l][r] = 1;
    }
  }
  for (int i = 0; i < lo; i++) {
    for (int j = 0; j <= i; j++) {
      sum[i] += p[j][i];
    }
  }
  for (int i = 1; i < lo; i++) {
    sum[i] += sum[i - 1];
  }
  long long ans = 0;
  for (int i = 0; i < lo; i++) {
    for (int j = 0; j <= i; j++) {
      if (p[j][i]) ans += sum[j - 1];
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 int maxn = 2000 + 5;
int p[maxn][maxn];
char ch[maxn];
int n;
long long dp[maxn];
long long sum[maxn];
int main() {
  scanf("%s", ch);
  int lo = 0;
  for (int i = 0; ch[i] != '\0'; i++) {
    lo++;
  }
  for (int i = 0; i < lo; i++) {
    p[i][i] = 1;
    for (int l = i - 1, r = i + 1; l >= 0 && r < lo; l--, r++) {
      if (ch[l] != ch[r]) break;
      p[l][r] = 1;
    }
    for (int l = i, r = i + 1; l >= 0 && r < lo; l--, r++) {
      if (ch[l] != ch[r]) break;
      p[l][r] = 1;
    }
  }
  for (int i = 0; i < lo; i++) {
    for (int j = 0; j <= i; j++) {
      sum[i] += p[j][i];
    }
  }
  for (int i = 1; i < lo; i++) {
    sum[i] += sum[i - 1];
  }
  long long ans = 0;
  for (int i = 0; i < lo; i++) {
    for (int j = 0; j <= i; j++) {
      if (p[j][i]) ans += sum[j - 1];
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```