#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int dp[301][18];
int solve(int amount, int num) {
  // cerr << "solve(" << amount << "," << num << ")" << endl;
  if(amount == 0) return 1;
  if(num < 0) return 0;
  int &res = dp[amount][num];
  if(res != -1) return res;
  res = 0;
  for(int k = num; k >= 1; --k) {
    if(amount - k*k >= 0) res += solve(amount - k*k, k);
  }
  return res;
}

int main() {
  memset(dp, -1, sizeof(dp));
  while(true) {
    int N; cin >> N;
    if(N == 0) break;
    cout << solve(N, 17) << endl;
  }
}
### Prompt

Please provide a cpp coded solution to the problem described below:

People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (= 172), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland.

There are four combinations of coins to pay ten credits:

* ten 1-credit coins,
* one 4-credit coin and six 1-credit coins,
* two 4-credit coins and two 1-credit coins, and
* one 9-credit coin and one 1-credit coin.



Your mission is to count the number of ways to pay a given amount using coins of Silverland.



Input

The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.

Output

For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output.

Example

Input

2
10
30
0


Output

1
4
27

### Response

```cpp
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int dp[301][18];
int solve(int amount, int num) {
  // cerr << "solve(" << amount << "," << num << ")" << endl;
  if(amount == 0) return 1;
  if(num < 0) return 0;
  int &res = dp[amount][num];
  if(res != -1) return res;
  res = 0;
  for(int k = num; k >= 1; --k) {
    if(amount - k*k >= 0) res += solve(amount - k*k, k);
  }
  return res;
}

int main() {
  memset(dp, -1, sizeof(dp));
  while(true) {
    int N; cin >> N;
    if(N == 0) break;
    cout << solve(N, 17) << endl;
  }
}
```