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


int square[] = { 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289 };
int ans;


void dfs(int num, int upper){
	//cout << num << "\t" << upper << endl;

	for (int i = upper; i >= 0; i--){
		if (num > square[i]) dfs(num - square[i], i);
		else if (num == square[i]) ans++;
	}

	return;
}


int main(void){

	int num;

	while (cin >> num, num){
		ans = 0;
		dfs(num, 16);

		cout << ans << endl;

	}

	return 0;
}
### 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<bits/stdc++.h>
using namespace std;


int square[] = { 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289 };
int ans;


void dfs(int num, int upper){
	//cout << num << "\t" << upper << endl;

	for (int i = upper; i >= 0; i--){
		if (num > square[i]) dfs(num - square[i], i);
		else if (num == square[i]) ans++;
	}

	return;
}


int main(void){

	int num;

	while (cin >> num, num){
		ans = 0;
		dfs(num, 16);

		cout << ans << endl;

	}

	return 0;
}
```