#include <bits/stdc++.h>
using namespace std;
char s[4010], ss[4010];
long long cnt[4010], tot[4010];
int dp[2010][2010];
int main() {
  int i, j, n, m, len, nlen;
  long long ans;
  while (scanf("%s", s) != EOF) {
    len = strlen(s);
    ans = 0;
    memset(dp, 0, sizeof(dp));
    memset(cnt, 0, sizeof(cnt));
    for (i = 0; i < len; i++) dp[i][i] = 1;
    for (i = 1; i < len; i++)
      if (s[i] == s[i - 1]) dp[i - 1][i] = 1;
    for (i = 2; i < len; i++) {
      for (j = 0; j + i < len; j++) {
        if (s[j + i] == s[j]) dp[j][j + i] = dp[j + 1][j + i - 1];
      }
    }
    long long tt;
    for (i = len - 1; i >= 0; i--) {
      tt = 0;
      for (j = i; j < len; j++) {
        if (dp[i][j]) tt++;
      }
      cnt[i] = tt + cnt[i + 1];
    }
    ans = 0;
    for (i = 0; i < len; i++) {
      for (j = i; j < len; j++) {
        if (dp[i][j]) ans += cnt[j + 1];
      }
    }
    printf("%I64d\n", ans);
  }
  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[4010], ss[4010];
long long cnt[4010], tot[4010];
int dp[2010][2010];
int main() {
  int i, j, n, m, len, nlen;
  long long ans;
  while (scanf("%s", s) != EOF) {
    len = strlen(s);
    ans = 0;
    memset(dp, 0, sizeof(dp));
    memset(cnt, 0, sizeof(cnt));
    for (i = 0; i < len; i++) dp[i][i] = 1;
    for (i = 1; i < len; i++)
      if (s[i] == s[i - 1]) dp[i - 1][i] = 1;
    for (i = 2; i < len; i++) {
      for (j = 0; j + i < len; j++) {
        if (s[j + i] == s[j]) dp[j][j + i] = dp[j + 1][j + i - 1];
      }
    }
    long long tt;
    for (i = len - 1; i >= 0; i--) {
      tt = 0;
      for (j = i; j < len; j++) {
        if (dp[i][j]) tt++;
      }
      cnt[i] = tt + cnt[i + 1];
    }
    ans = 0;
    for (i = 0; i < len; i++) {
      for (j = i; j < len; j++) {
        if (dp[i][j]) ans += cnt[j + 1];
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```