#include <bits/stdc++.h>
std::vector<long long> values;
std::vector<long long> left;
std::vector<std::vector<bool>> palindrom;
std::string line;
std::vector<int> entry[26];
bool ispalindrom(int start, int finish) {
  for (int i = 0; i <= (finish - start) / 2; i++) {
    if (line[start + i] != line[finish - i]) return false;
  }
  return true;
}
int main() {
  std::cin >> line;
  values.resize(line.length());
  left.resize(line.length());
  palindrom.resize(line.length());
  for (int i = 0; i < line.length(); i++) palindrom[i].resize(line.length());
  for (int j = 0; j < line.length(); j++) {
    palindrom[j][j] = true;
    left[j]++;
    int i = 1;
    while ((j + i < line.length()) && (j - i >= 0)) {
      if (line[j + i] == line[j - i]) {
        palindrom[j - i][j + i] = true;
        left[j + i]++;
      } else {
        break;
      };
      i++;
    }
  }
  for (int j = 0; j < line.length() - 1; j++) {
    int i = 0;
    if (line[j] != line[j + 1]) continue;
    while ((j - i >= 0) && (j + 1 + i < line.length())) {
      if (line[j - i] == line[j + 1 + i]) {
        palindrom[j - i][j + 1 + i] = true;
        left[j + 1 + i]++;
      } else {
        break;
      }
      i++;
    }
  }
  for (int j = 1; j < line.length(); j++) {
    left[j] += left[j - 1];
  }
  values[0] = 0;
  values[1] = 1;
  for (int i = 2; i < line.length(); i++) {
    values[i] = values[i - 1];
    for (int j = 1; j <= i; j++) {
      if (palindrom[j][i]) {
        values[i] += left[j - 1];
      }
    }
  }
  std::cout << values[line.length() - 1];
}

### Prompt

Please formulate 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>
std::vector<long long> values;
std::vector<long long> left;
std::vector<std::vector<bool>> palindrom;
std::string line;
std::vector<int> entry[26];
bool ispalindrom(int start, int finish) {
  for (int i = 0; i <= (finish - start) / 2; i++) {
    if (line[start + i] != line[finish - i]) return false;
  }
  return true;
}
int main() {
  std::cin >> line;
  values.resize(line.length());
  left.resize(line.length());
  palindrom.resize(line.length());
  for (int i = 0; i < line.length(); i++) palindrom[i].resize(line.length());
  for (int j = 0; j < line.length(); j++) {
    palindrom[j][j] = true;
    left[j]++;
    int i = 1;
    while ((j + i < line.length()) && (j - i >= 0)) {
      if (line[j + i] == line[j - i]) {
        palindrom[j - i][j + i] = true;
        left[j + i]++;
      } else {
        break;
      };
      i++;
    }
  }
  for (int j = 0; j < line.length() - 1; j++) {
    int i = 0;
    if (line[j] != line[j + 1]) continue;
    while ((j - i >= 0) && (j + 1 + i < line.length())) {
      if (line[j - i] == line[j + 1 + i]) {
        palindrom[j - i][j + 1 + i] = true;
        left[j + 1 + i]++;
      } else {
        break;
      }
      i++;
    }
  }
  for (int j = 1; j < line.length(); j++) {
    left[j] += left[j - 1];
  }
  values[0] = 0;
  values[1] = 1;
  for (int i = 2; i < line.length(); i++) {
    values[i] = values[i - 1];
    for (int j = 1; j <= i; j++) {
      if (palindrom[j][i]) {
        values[i] += left[j - 1];
      }
    }
  }
  std::cout << values[line.length() - 1];
}

```