#include <bits/stdc++.h>
using namespace std;
int n;
char s[2000];
long long dp[2000];
int isp[2000][2000];
bool isPal(int i, int j, int type) {
  int b, e;
  if (type == 1)
    b = i - j, e = i + j;
  else
    b = i - j + 1;
  e = i + j;
  if (isp[b][e] >= 0) return (bool)isp[b][e];
  if (b == e) return isp[b][e] = true;
  if (b + 1 == e) return isp[b][e] = (s[b] == s[e]);
  if (isp[b + 1][e - 1])
    return isp[b][e] = (s[b] == s[e]);
  else
    return isp[b][e] = false;
}
long long getNum() {
  int len = strlen(s), i, j;
  bool r;
  memset(isp, -1, 2000 * 2000 * sizeof(int));
  for (i = 0; i < len; i++)
    for (j = 0; j <= len / 2; j++)
      if (i + j < len) {
        if (i - j >= 0) {
          r = isPal(i, j, 1);
          if (0) printf("(%d %d %d): %d\n", i, j, 1, r);
        }
        if (j > 0 && i - j + 1 >= 0) {
          r = isPal(i, j, 0);
          if (0) printf("(%d %d %d): %d\n", i, j, 0, r);
        }
      }
  dp[len - 1] = 1;
  for (i = len - 2; i >= 0; i--) {
    dp[i] = dp[i + 1];
    for (j = i; j < len; j++) dp[i] += isp[i][j];
  }
  long long sum = 0;
  for (i = 0; i < len; i++)
    for (j = i; j < len - 1; j++)
      if (isp[i][j]) sum += dp[j + 1];
  return sum;
}
int main() {
  scanf("%s", s);
  printf("%I64d", getNum());
  return 0;
}

### Prompt

Generate 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;
int n;
char s[2000];
long long dp[2000];
int isp[2000][2000];
bool isPal(int i, int j, int type) {
  int b, e;
  if (type == 1)
    b = i - j, e = i + j;
  else
    b = i - j + 1;
  e = i + j;
  if (isp[b][e] >= 0) return (bool)isp[b][e];
  if (b == e) return isp[b][e] = true;
  if (b + 1 == e) return isp[b][e] = (s[b] == s[e]);
  if (isp[b + 1][e - 1])
    return isp[b][e] = (s[b] == s[e]);
  else
    return isp[b][e] = false;
}
long long getNum() {
  int len = strlen(s), i, j;
  bool r;
  memset(isp, -1, 2000 * 2000 * sizeof(int));
  for (i = 0; i < len; i++)
    for (j = 0; j <= len / 2; j++)
      if (i + j < len) {
        if (i - j >= 0) {
          r = isPal(i, j, 1);
          if (0) printf("(%d %d %d): %d\n", i, j, 1, r);
        }
        if (j > 0 && i - j + 1 >= 0) {
          r = isPal(i, j, 0);
          if (0) printf("(%d %d %d): %d\n", i, j, 0, r);
        }
      }
  dp[len - 1] = 1;
  for (i = len - 2; i >= 0; i--) {
    dp[i] = dp[i + 1];
    for (j = i; j < len; j++) dp[i] += isp[i][j];
  }
  long long sum = 0;
  for (i = 0; i < len; i++)
    for (j = i; j < len - 1; j++)
      if (isp[i][j]) sum += dp[j + 1];
  return sum;
}
int main() {
  scanf("%s", s);
  printf("%I64d", getNum());
  return 0;
}

```