#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<map>
#include<cstring>
using namespace std;

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(n) rep(i,n)
#define all(n) n.begin(),n.end()

const int MAXC = 17, MAXN = 301;

int c[MAXC];

int dp[MAXC + 1][MAXN];

int main()
{
	REP(MAXC) c[i] = (i+1) * (i+1);

	dp[0][0] = 1;
	REP(MAXC)rep(j,MAXN)for(int w = 0;w < MAXN;w++)
	{
		if(j + c[i] * w >= MAXN) break;

		dp[i+1][j + c[i] * w] += dp[i][j];
	}

	int	n;
	while(cin >> n && n)
	{
		cout << dp[MAXC][n] << endl;
	}

	return 0;
}
### Prompt

Develop a solution in Cpp 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<string>
#include<algorithm>
#include<vector>
#include<map>
#include<cstring>
using namespace std;

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(n) rep(i,n)
#define all(n) n.begin(),n.end()

const int MAXC = 17, MAXN = 301;

int c[MAXC];

int dp[MAXC + 1][MAXN];

int main()
{
	REP(MAXC) c[i] = (i+1) * (i+1);

	dp[0][0] = 1;
	REP(MAXC)rep(j,MAXN)for(int w = 0;w < MAXN;w++)
	{
		if(j + c[i] * w >= MAXN) break;

		dp[i+1][j + c[i] * w] += dp[i][j];
	}

	int	n;
	while(cin >> n && n)
	{
		cout << dp[MAXC][n] << endl;
	}

	return 0;
}
```