#include<iostream>
#include<cstdio>

using namespace std;

int coins[]={0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289};

int main()
{
	int m;

	while (scanf("%d", &m))
	{
		if (m==0)
		break;

		long long int T[18][m+1];

		for (int i=0; i<18; i++)
		{
			for (int j=0; j<=m; j++)
			{
				if (i==0)
				T[i][j]=0;
				else if (j==0)
				T[i][j]=1;
				else if (j>=coins[i])
				T[i][j]=T[i-1][j]+T[i][j-coins[i]];
				else
				T[i][j]=T[i-1][j];
			}
		}

		printf("%d\n",T[17][m]);
	}

	return 0;
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

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<cstdio>

using namespace std;

int coins[]={0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289};

int main()
{
	int m;

	while (scanf("%d", &m))
	{
		if (m==0)
		break;

		long long int T[18][m+1];

		for (int i=0; i<18; i++)
		{
			for (int j=0; j<=m; j++)
			{
				if (i==0)
				T[i][j]=0;
				else if (j==0)
				T[i][j]=1;
				else if (j>=coins[i])
				T[i][j]=T[i-1][j]+T[i][j-coins[i]];
				else
				T[i][j]=T[i-1][j];
			}
		}

		printf("%d\n",T[17][m]);
	}

	return 0;
}
```