#include<bits/stdc++.h>
using namespace std;

int count( int S[], int m, int n )
{
    if (n == 0)
        return 1;
    if (n < 0)
        return 0;
    if (m <=0 && n >= 1)
        return 0;
    return count( S, m - 1, n ) + count( S, m, n-S[m-1] );
}
int main()
{
    int n;
    while(cin>>n){
     if(n==0) exit(0);
     else{
    int arr[300];
     int j=0;
     for(int i=1;i<=n;i++){
        if((i*i)<=n){
            arr[j++]=i*i;
        }
     }
    cout<<count(arr,j,n)<<endl;
    }
    }
    return 0;
}
### Prompt

Create a solution in cpp for 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<bits/stdc++.h>
using namespace std;

int count( int S[], int m, int n )
{
    if (n == 0)
        return 1;
    if (n < 0)
        return 0;
    if (m <=0 && n >= 1)
        return 0;
    return count( S, m - 1, n ) + count( S, m, n-S[m-1] );
}
int main()
{
    int n;
    while(cin>>n){
     if(n==0) exit(0);
     else{
    int arr[300];
     int j=0;
     for(int i=1;i<=n;i++){
        if((i*i)<=n){
            arr[j++]=i*i;
        }
     }
    cout<<count(arr,j,n)<<endl;
    }
    }
    return 0;
}
```