#include <bits/stdc++.h>
using namespace std;
char s[2005];
long long len, dp[2005][2005], dp1[20005], kq;
void khoitri() {
  long long i;
  for (i = 0; i < len; i++) {
    dp[i][i] = 1;
    if (i != len - 1 && s[i] == s[i + 1]) dp[i][i + 1] = 1;
  }
}
void solve() {
  long long i, j, dau, cuoi;
  len = strlen(s);
  khoitri();
  for (i = 2; i < len; i++) {
    for (j = 0; j <= len - i; j++) {
      dau = j;
      cuoi = j + i;
      if (s[dau] == s[cuoi]) {
        dp[dau][cuoi] = dp[dau + 1][cuoi - 1];
      } else {
        dp[dau][cuoi] = 0;
      }
    }
  }
  for (i = 0; i < len; i++) {
    dp1[i] = 0;
    for (j = i; j < len; j++) {
      dp1[i] += dp[i][j];
    }
  }
  for (i = len - 2; i >= 0; i--) dp1[i] += dp1[i + 1];
  kq = 0;
  for (i = 0; i < len - 1; i++) {
    for (j = i; j < len - 1; j++) {
      if (dp[i][j] != 0) {
        kq += dp1[j + 1];
      }
    }
  }
  printf("%I64d\n", kq);
}
int main() {
  while (scanf("%s", &s) > 0) {
    solve();
  }
  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>
using namespace std;
char s[2005];
long long len, dp[2005][2005], dp1[20005], kq;
void khoitri() {
  long long i;
  for (i = 0; i < len; i++) {
    dp[i][i] = 1;
    if (i != len - 1 && s[i] == s[i + 1]) dp[i][i + 1] = 1;
  }
}
void solve() {
  long long i, j, dau, cuoi;
  len = strlen(s);
  khoitri();
  for (i = 2; i < len; i++) {
    for (j = 0; j <= len - i; j++) {
      dau = j;
      cuoi = j + i;
      if (s[dau] == s[cuoi]) {
        dp[dau][cuoi] = dp[dau + 1][cuoi - 1];
      } else {
        dp[dau][cuoi] = 0;
      }
    }
  }
  for (i = 0; i < len; i++) {
    dp1[i] = 0;
    for (j = i; j < len; j++) {
      dp1[i] += dp[i][j];
    }
  }
  for (i = len - 2; i >= 0; i--) dp1[i] += dp1[i + 1];
  kq = 0;
  for (i = 0; i < len - 1; i++) {
    for (j = i; j < len - 1; j++) {
      if (dp[i][j] != 0) {
        kq += dp1[j + 1];
      }
    }
  }
  printf("%I64d\n", kq);
}
int main() {
  while (scanf("%s", &s) > 0) {
    solve();
  }
  return 0;
}

```