#include <iostream>

using namespace std;
 
int a[305];
int b[305];

int main() {
    int n;
    
    while(cin >> n && n != 0) {
    	
        for(int i = 0; i <= n; i++) {
        	
            a[i] = 1;
            b[i] = 0;
            
        }
        
        for(int i = 2; i <= 17; i++) {
        	
            for(int j = 0; j <= n; j++) 
            	for(int k = 0; k+j <= n; k+=i*i)
					b[j+k] += a[j];
					
            for(int j = 0; j <= n; j++) {
            	
                	a[j] = b[j];
					b[j] = 0;
					
            }
            
        }
        
        cout << a[n] << "\n";
        
    }
    
    return 0;
}

### Prompt

Your task is to create 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>

using namespace std;
 
int a[305];
int b[305];

int main() {
    int n;
    
    while(cin >> n && n != 0) {
    	
        for(int i = 0; i <= n; i++) {
        	
            a[i] = 1;
            b[i] = 0;
            
        }
        
        for(int i = 2; i <= 17; i++) {
        	
            for(int j = 0; j <= n; j++) 
            	for(int k = 0; k+j <= n; k+=i*i)
					b[j+k] += a[j];
					
            for(int j = 0; j <= n; j++) {
            	
                	a[j] = b[j];
					b[j] = 0;
					
            }
            
        }
        
        cout << a[n] << "\n";
        
    }
    
    return 0;
}

```