#include<iostream>
using namespace std;

int main(){
  //set;
  int a[17];
  for(int i=0;i<17;i++){
    a[i]=(i+1)*(i+1);
  }
  
  int data[17][301]={0};
  for(int i=0;i<301;i++)data[0][i]=1;
  data[0][0]=0;
  for(int i=1;i<17;i++){
    data[i][0]=0;
    for(int j=1;j<301;j++){
      if (a[i] >j){
        data[i][j]=data[i-1][j];
      }else if (a[i] == j)data[i][j]=data[i-1][j]+1;
      else {
        data[i][j]=data[i-1][j]+data[i][j-a[i]];
      }
    }
  }
  
  int n;
  while(cin >> n&&n)cout <<data[16][n]<<endl;
  
  
}
### Prompt

Your challenge is to write 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<iostream>
using namespace std;

int main(){
  //set;
  int a[17];
  for(int i=0;i<17;i++){
    a[i]=(i+1)*(i+1);
  }
  
  int data[17][301]={0};
  for(int i=0;i<301;i++)data[0][i]=1;
  data[0][0]=0;
  for(int i=1;i<17;i++){
    data[i][0]=0;
    for(int j=1;j<301;j++){
      if (a[i] >j){
        data[i][j]=data[i-1][j];
      }else if (a[i] == j)data[i][j]=data[i-1][j]+1;
      else {
        data[i][j]=data[i-1][j]+data[i][j-a[i]];
      }
    }
  }
  
  int n;
  while(cin >> n&&n)cout <<data[16][n]<<endl;
  
  
}
```