#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
using namespace std;

int dp[301][18];

int main(){
	int n;
	while(cin >> n, n){
		for(int i = 0; i <= n; i++) dp[i][1] = dp[i][0] = 1;
		for(int i = 0; i <= n; i++){
			for(int j = 2; j <= 18; j++){
				dp[i][j] = 0;
				int k;
				for(k = 0; i - k * j * j >= 0; k++){
					dp[i][j] += dp[i - k * j * j][j - 1];
				}
			}
		}
		cout << dp[n][18] << endl;
	}
}
### Prompt

In CPP, your task is to solve 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 <vector>
#include <string>
#include <map>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
using namespace std;

int dp[301][18];

int main(){
	int n;
	while(cin >> n, n){
		for(int i = 0; i <= n; i++) dp[i][1] = dp[i][0] = 1;
		for(int i = 0; i <= n; i++){
			for(int j = 2; j <= 18; j++){
				dp[i][j] = 0;
				int k;
				for(k = 0; i - k * j * j >= 0; k++){
					dp[i][j] += dp[i - k * j * j][j - 1];
				}
			}
		}
		cout << dp[n][18] << endl;
	}
}
```