#include <iostream>
#include <vector>
using namespace std;
int main(){
	vector<int> tbl;
	int dp[301] = {0};
	
	
	for(int i=1 ; i*i<= 300 ; i++)tbl.push_back(i*i);
	
	dp[0] = 1;
	
	for(int j = 0; j < tbl.size() ; j++){
		for(int co = tbl[j]; co <= 300 ; co++){
				dp[co] += dp[co-tbl[j]];
		}
	}
	int n;
	while(cin >> n , n){
		cout << dp[n] << endl;
	}
}

	
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
using namespace std;
int main(){
	vector<int> tbl;
	int dp[301] = {0};
	
	
	for(int i=1 ; i*i<= 300 ; i++)tbl.push_back(i*i);
	
	dp[0] = 1;
	
	for(int j = 0; j < tbl.size() ; j++){
		for(int co = tbl[j]; co <= 300 ; co++){
				dp[co] += dp[co-tbl[j]];
		}
	}
	int n;
	while(cin >> n , n){
		cout << dp[n] << endl;
	}
}

	
```