#include<iostream>
using namespace std;

int main()
{
  
  int coins[17];

  for(int i = 0; i < 17; ++i){
    coins[i] = (i+1)*(i+1);
  }

  long long dp[300][17] = {{0,}};

  for(int i = 0; i < 17; ++i){
    dp[coins[i]][i] = 1;
  }
  for(int i = 0; i < 300; ++i){
    for(int j = 0; j < 17; ++j){
      for(int k = j; k < 17; ++k){
        if(i+coins[k]<300){
          dp[i+coins[k]][k] += dp[i][j];
        }
      }
    }
  }

  int p;
  while(cin>>p){
    if(p==0)break;
    long long ret = 0;
    for(int i = 0; i < 17; ++i){
      ret += dp[p][i];
    }
    cout << ret << endl;
  }
  return 0;
}
### Prompt

Please provide a Cpp coded solution 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>
using namespace std;

int main()
{
  
  int coins[17];

  for(int i = 0; i < 17; ++i){
    coins[i] = (i+1)*(i+1);
  }

  long long dp[300][17] = {{0,}};

  for(int i = 0; i < 17; ++i){
    dp[coins[i]][i] = 1;
  }
  for(int i = 0; i < 300; ++i){
    for(int j = 0; j < 17; ++j){
      for(int k = j; k < 17; ++k){
        if(i+coins[k]<300){
          dp[i+coins[k]][k] += dp[i][j];
        }
      }
    }
  }

  int p;
  while(cin>>p){
    if(p==0)break;
    long long ret = 0;
    for(int i = 0; i < 17; ++i){
      ret += dp[p][i];
    }
    cout << ret << endl;
  }
  return 0;
}
```