#include <iostream>
#include <vector>
#include <cstring>
using namespace std;

#define SIZE 300
#define INDEX_END 17

int c[SIZE + 1][INDEX_END + 1];

int run(int n, int index) {
    if (c[n][index] > 0) {
        return c[n][index];
    }
    if (n == 0) {
        return 1;
    }
    if (INDEX_END < index || n < index * index) {
        return 0;
    }
    int result = 0;
    result += run(n, index + 1);
    result += run(n - index * index, index);
    return c[n][index] = result;
}

int main() {
    memset(c, 0, sizeof(c));
    while (true) {
        int n;
        cin >> n;
        if (n == 0) {
            break;
        }
        cout << run(n, 1) << endl;
    }
}
### 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 <iostream>
#include <vector>
#include <cstring>
using namespace std;

#define SIZE 300
#define INDEX_END 17

int c[SIZE + 1][INDEX_END + 1];

int run(int n, int index) {
    if (c[n][index] > 0) {
        return c[n][index];
    }
    if (n == 0) {
        return 1;
    }
    if (INDEX_END < index || n < index * index) {
        return 0;
    }
    int result = 0;
    result += run(n, index + 1);
    result += run(n - index * index, index);
    return c[n][index] = result;
}

int main() {
    memset(c, 0, sizeof(c));
    while (true) {
        int n;
        cin >> n;
        if (n == 0) {
            break;
        }
        cout << run(n, 1) << endl;
    }
}
```