#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <complex>
using namespace std;

#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define all(x) (x).begin(),(x).end()
#define mp make_pair
#define pb push_back
#define EPS 1e-8
#define DEB 0

int dp[310];
int coin[32];

int main(){
	int n;
	
	rep(i,17){
		coin[i] = (i+1)*(i+1);
	}
#if DEB
	rep(i,17){
		printf("%d:%d\n",i,coin[i]);
	}
#endif	

	dp[0] = 1;
	rep(i,17){
		for(int j=coin[i]; j<=300; j++){
			dp[j] += dp[j-coin[i]];
		}
	}
	
	while(scanf("%d",&n),n){
		printf("%d\n",dp[n]);
	}
	return 0;
}
### Prompt

Your challenge is to write 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 <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <complex>
using namespace std;

#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define all(x) (x).begin(),(x).end()
#define mp make_pair
#define pb push_back
#define EPS 1e-8
#define DEB 0

int dp[310];
int coin[32];

int main(){
	int n;
	
	rep(i,17){
		coin[i] = (i+1)*(i+1);
	}
#if DEB
	rep(i,17){
		printf("%d:%d\n",i,coin[i]);
	}
#endif	

	dp[0] = 1;
	rep(i,17){
		for(int j=coin[i]; j<=300; j++){
			dp[j] += dp[j-coin[i]];
		}
	}
	
	while(scanf("%d",&n),n){
		printf("%d\n",dp[n]);
	}
	return 0;
}
```