#include<iostream>
using namespace std;
int main()
{
  int n,c[17],p[300]={0},i,j;
  for(i=1;i<18;i++)c[i-1]=i*i;
  p[0]=1;
  for(i=0;i<17;i++)
    for(j=0;j<300;j++)
      if(j-c[i]>=0)p[j]+=p[j-c[i]];
  while(cin>>n,n)cout<<p[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()
{
  int n,c[17],p[300]={0},i,j;
  for(i=1;i<18;i++)c[i-1]=i*i;
  p[0]=1;
  for(i=0;i<17;i++)
    for(j=0;j<300;j++)
      if(j-c[i]>=0)p[j]+=p[j-c[i]];
  while(cin>>n,n)cout<<p[n]<<endl;
}
```