#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>

#define REP(i,n) for(long long i=0;i<n;++i)
#define REPR(i,n) for(long long i=n;i>=0;--i)
#define REPI(itr,v) for(auto itr=v.begin();itr!=v.end();++itr)
#define REPIR(itr,v) for(auto itr=v.rbegin();itr!=v.rend();++itr)
#define FOR(i,a,b) for(long long i=a;i<b;++i)
#define SORT(v,n) sort(v, v+n)
#define SORTV(v) sort(v.begin(), v.end())
#define ALL(v) v.begin(),v.end()
#define llong long long
#define INF 999999999
#define SUR 1000000007
#define pb push_back
#define pf push_front
#define MP make_pair
#define SV(v) {for(long long sitr=0;sitr<v.size();++sitr){cin>>v[sitr];}}

int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};

using namespace std;

typedef pair<int,int> pii;

llong solve(int n, int coin){
  if(coin == 1 || n == 0) return 1;

  llong res = 0;
  int c = coin * coin;
  for(int i = 0; n - i * c >= 0; ++i){
    res += solve(n - i * c, coin-1);
  }
  return res;
}

int main(){

  while(true){
    int n;
    cin >> n;
    if(n == 0) break;

    cout << solve(n, 17) << endl;
  }
  return 0;
}



### Prompt

Please formulate 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 <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>

#define REP(i,n) for(long long i=0;i<n;++i)
#define REPR(i,n) for(long long i=n;i>=0;--i)
#define REPI(itr,v) for(auto itr=v.begin();itr!=v.end();++itr)
#define REPIR(itr,v) for(auto itr=v.rbegin();itr!=v.rend();++itr)
#define FOR(i,a,b) for(long long i=a;i<b;++i)
#define SORT(v,n) sort(v, v+n)
#define SORTV(v) sort(v.begin(), v.end())
#define ALL(v) v.begin(),v.end()
#define llong long long
#define INF 999999999
#define SUR 1000000007
#define pb push_back
#define pf push_front
#define MP make_pair
#define SV(v) {for(long long sitr=0;sitr<v.size();++sitr){cin>>v[sitr];}}

int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};

using namespace std;

typedef pair<int,int> pii;

llong solve(int n, int coin){
  if(coin == 1 || n == 0) return 1;

  llong res = 0;
  int c = coin * coin;
  for(int i = 0; n - i * c >= 0; ++i){
    res += solve(n - i * c, coin-1);
  }
  return res;
}

int main(){

  while(true){
    int n;
    cin >> n;
    if(n == 0) break;

    cout << solve(n, 17) << endl;
  }
  return 0;
}



```