#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <complex>
#include <queue>
#include <map>
#include <set>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cmath>
using namespace std;

#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
const int INF = 1<<29;

int main() {
  int dp[301];
  memset(dp,0,sizeof(dp));
  dp[0] = 1;
  vector<int> coin;
  for (int i=1; i*i<=289; ++i)
    coin.push_back(i*i);
  FOR(it, coin) {
    REP(i,300) {
      if (i+*it>300) break;
      dp[i+*it] += dp[i];
    }
  }
  int n;
  while(cin >> n, n)
    cout << dp[n] << 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 <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <complex>
#include <queue>
#include <map>
#include <set>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cmath>
using namespace std;

#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
const int INF = 1<<29;

int main() {
  int dp[301];
  memset(dp,0,sizeof(dp));
  dp[0] = 1;
  vector<int> coin;
  for (int i=1; i*i<=289; ++i)
    coin.push_back(i*i);
  FOR(it, coin) {
    REP(i,300) {
      if (i+*it>300) break;
      dp[i+*it] += dp[i];
    }
  }
  int n;
  while(cin >> n, n)
    cout << dp[n] << endl;
}
```