#include<stdio.h>
#include<iostream>
#include<string>
#include<algorithm>
#include <vector>
#include<math.h>
#include<iomanip>


using namespace std;

int main(){
	int X[312]={};
	for(int i=0;i<=300;i++)
		X[i]=1;
	/*
	for(int i=0;i<301;i++){
		X[i+4]+=X[i];if(i>=8)X[i+4]--;
		X[i+9]+=X[i];if(i>=18)X[i+9]--;

	}
	*/
	for(int i=2;i<=17;i++)
	{
		int p=i*i;X[p]++;
		for(int j=1;j+p<301;j++)
		{
			X[j+p]+=X[j];//if(j>=2*p)X[j+p]--;
		}
	}
	/*
	X[0]=1;
	for(int i=1;i<301;i++)
		X[i]+=X[i-1];

		*/

	int d;
	while(1){
		cin>>d;
		if(d==0)break;
		cout<<X[d]<<endl;
	}
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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<stdio.h>
#include<iostream>
#include<string>
#include<algorithm>
#include <vector>
#include<math.h>
#include<iomanip>


using namespace std;

int main(){
	int X[312]={};
	for(int i=0;i<=300;i++)
		X[i]=1;
	/*
	for(int i=0;i<301;i++){
		X[i+4]+=X[i];if(i>=8)X[i+4]--;
		X[i+9]+=X[i];if(i>=18)X[i+9]--;

	}
	*/
	for(int i=2;i<=17;i++)
	{
		int p=i*i;X[p]++;
		for(int j=1;j+p<301;j++)
		{
			X[j+p]+=X[j];//if(j>=2*p)X[j+p]--;
		}
	}
	/*
	X[0]=1;
	for(int i=1;i<301;i++)
		X[i]+=X[i-1];

		*/

	int d;
	while(1){
		cin>>d;
		if(d==0)break;
		cout<<X[d]<<endl;
	}
	return 0;
}
```