#include <iostream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;

int main()
{
    vector<int> num(301, 0);
    num[0] = 1;
    for(int i=1; i<=17; ++i){
        for(int j=0; j<300; ++j){
            int k = i*i + j;
            if(k <= 300)
                num[k] += num[j];
        }
    }

    for(;;){
        int n;
        cin >> n;
        if(n == 0)
            return 0;
        cout << num[n] << endl;
    }
}
### Prompt

Please formulate 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 <sstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;

int main()
{
    vector<int> num(301, 0);
    num[0] = 1;
    for(int i=1; i<=17; ++i){
        for(int j=0; j<300; ++j){
            int k = i*i + j;
            if(k <= 300)
                num[k] += num[j];
        }
    }

    for(;;){
        int n;
        cin >> n;
        if(n == 0)
            return 0;
        cout << num[n] << endl;
    }
}
```