#include <iostream>

using namespace std;

int coin[] = {289, 256, 225, 196, 169, 144, 121, 100, 81, 64, 49, 36, 25, 16, 9, 4, 1};
   
int solve(int pay, int n) {
    if (n == 16) return 1;
    if (pay < 4) return 1;
    
    int sum = 0;
    int no_can_use = pay / coin[n];
    if (no_can_use == 0) return solve(pay, n+1);
    for (int i = 0; i <= no_can_use; i++) {
        sum += solve(pay - coin[n] * i, n+1);
    }

    return sum;
}

int main() {
    int n;
    while (cin >> n) {
        if (n == 0) break;
        cout << solve(n, 0) << 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 <iostream>

using namespace std;

int coin[] = {289, 256, 225, 196, 169, 144, 121, 100, 81, 64, 49, 36, 25, 16, 9, 4, 1};
   
int solve(int pay, int n) {
    if (n == 16) return 1;
    if (pay < 4) return 1;
    
    int sum = 0;
    int no_can_use = pay / coin[n];
    if (no_can_use == 0) return solve(pay, n+1);
    for (int i = 0; i <= no_can_use; i++) {
        sum += solve(pay - coin[n] * i, n+1);
    }

    return sum;
}

int main() {
    int n;
    while (cin >> n) {
        if (n == 0) break;
        cout << solve(n, 0) << endl;
    }
    return 0;
}
```