#include <iostream>
#define N 17
#define N1 301
using namespace std;
void DP(int);
int main(){
  int n;
  while(1){
    cin >> n;
    if(n==0) break;
    DP(n);
  }
  return 0;
}
void DP(int n){
  int dp[N][N1],c[N];
  for(int i=0;i<N;i++) dp[i][0]=1,c[i]=(i+1)*(i+1);
  for(int i=0;i<=n;i++) dp[0][i]=1;
  for(int i=1;i<N;i++){
    for(int j=1;j<=n;j++){
      if(c[i]<=j){
	dp[i][j]=dp[i-1][j]+dp[i][j-c[i]];
      }else{
	dp[i][j]=dp[i-1][j];
      }
    }
  }
  cout << dp[N-1][n] << endl;
}
### 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>
#define N 17
#define N1 301
using namespace std;
void DP(int);
int main(){
  int n;
  while(1){
    cin >> n;
    if(n==0) break;
    DP(n);
  }
  return 0;
}
void DP(int n){
  int dp[N][N1],c[N];
  for(int i=0;i<N;i++) dp[i][0]=1,c[i]=(i+1)*(i+1);
  for(int i=0;i<=n;i++) dp[0][i]=1;
  for(int i=1;i<N;i++){
    for(int j=1;j<=n;j++){
      if(c[i]<=j){
	dp[i][j]=dp[i-1][j]+dp[i][j-c[i]];
      }else{
	dp[i][j]=dp[i-1][j];
      }
    }
  }
  cout << dp[N-1][n] << endl;
}
```