#include <vector>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <functional>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <cassert>
using namespace std;
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define EACH(t,i,c) for(t::iterator i=(c).begin(); i!=(c).end(); ++i)
const double EPS = 1e-10;
const double PI  = acos(-1.0);

vi coins;

int memo[310][300]={};
int dfs(int n,int prev){
	if(memo[n][prev]){
		return memo[n][prev];
	}
	if(n==0){
		return 1;
	}
	int ret=0;
	REP(i,coins.size()){
		if(coins[i]<prev){
			continue;
		}
		if(coins[i]>n){
			break;
		}
		ret+=dfs(n-coins[i],coins[i]);
	}
	return memo[n][prev]=ret;
}
int main(){
	int n;
	int value=1;
	while(value*value<=300){
		coins.push_back(value*value);
		value++;
	}
	while(cin>>n,n){
		cout<<dfs(n,0)<<endl;
	}
}
### Prompt

Develop a solution in Cpp 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 <vector>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <functional>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <cassert>
using namespace std;
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define EACH(t,i,c) for(t::iterator i=(c).begin(); i!=(c).end(); ++i)
const double EPS = 1e-10;
const double PI  = acos(-1.0);

vi coins;

int memo[310][300]={};
int dfs(int n,int prev){
	if(memo[n][prev]){
		return memo[n][prev];
	}
	if(n==0){
		return 1;
	}
	int ret=0;
	REP(i,coins.size()){
		if(coins[i]<prev){
			continue;
		}
		if(coins[i]>n){
			break;
		}
		ret+=dfs(n-coins[i],coins[i]);
	}
	return memo[n][prev]=ret;
}
int main(){
	int n;
	int value=1;
	while(value*value<=300){
		coins.push_back(value*value);
		value++;
	}
	while(cin>>n,n){
		cout<<dfs(n,0)<<endl;
	}
}
```