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

int n;
long dp[290][18];

long rec(int a, int b){ //今の合計a,前に足した数b
	if(dp[a][b]>=0) return dp[a][b];
	
	long ret=0;	
	
	if(a==n) ret=1;
	else{
		for(int i=b; a+i*i<=n; ++i)	ret+=rec(a+i*i, i);
	}
	
	return dp[a][b]=ret;
}

int main(){
	while(1){
		scanf(" %d", &n);
		if(n==0) break;	
		
		memset(dp, -1, sizeof(dp));
		printf("%ld\n", rec(0,1));
	}
}
### Prompt

Your task is to create a Cpp solution to 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 <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n;
long dp[290][18];

long rec(int a, int b){ //今の合計a,前に足した数b
	if(dp[a][b]>=0) return dp[a][b];
	
	long ret=0;	
	
	if(a==n) ret=1;
	else{
		for(int i=b; a+i*i<=n; ++i)	ret+=rec(a+i*i, i);
	}
	
	return dp[a][b]=ret;
}

int main(){
	while(1){
		scanf(" %d", &n);
		if(n==0) break;	
		
		memset(dp, -1, sizeof(dp));
		printf("%ld\n", rec(0,1));
	}
}
```