#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <functional>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <tuple>
#include <complex>
#include <cassert>
#include <exception>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<P> vp;
#define rep(i,a,n) for(ll i = (a);i < (n);i++)
#define per(i,a,n) for(ll i = (a);i > (n);i--)
#define lep(i,a,n) for(ll i = (a);i <= (n);i++)
#define pel(i,a,n) for(ll i = (a);i >= (n);i--)
#define clr(a,b) memset((a),(b),sizeof(a))
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(),(c).end()
#define sz size()
#define print(X) cout << (X) << "\n"
static const int INF = 1e+9+7;
ll n,m,l;
string s,t;
ll d[200010],dp[3000][3000];
char field[200][200];

ll dfs(int x,int y){
  if(y == 1)return 1;
  ll ans = 0;
  for(int i = 0;i <= x;i += y * y)ans += dfs(x-i,y-1);
  return ans;
}

int main(){
  while(cin >> n,n){
    print(dfs(n,17));
  }
  return 0;
}

### 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 <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <functional>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <tuple>
#include <complex>
#include <cassert>
#include <exception>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<P> vp;
#define rep(i,a,n) for(ll i = (a);i < (n);i++)
#define per(i,a,n) for(ll i = (a);i > (n);i--)
#define lep(i,a,n) for(ll i = (a);i <= (n);i++)
#define pel(i,a,n) for(ll i = (a);i >= (n);i--)
#define clr(a,b) memset((a),(b),sizeof(a))
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(),(c).end()
#define sz size()
#define print(X) cout << (X) << "\n"
static const int INF = 1e+9+7;
ll n,m,l;
string s,t;
ll d[200010],dp[3000][3000];
char field[200][200];

ll dfs(int x,int y){
  if(y == 1)return 1;
  ll ans = 0;
  for(int i = 0;i <= x;i += y * y)ans += dfs(x-i,y-1);
  return ans;
}

int main(){
  while(cin >> n,n){
    print(dfs(n,17));
  }
  return 0;
}

```