#include <iostream>

using namespace std;

long long dp[18][301];

int main() {
  
  for(int i=1; i<=17; i++) {
    for(int j=0; j<=300; j++) {
      if(i == 1) {
	if(j == 0) dp[1][0] = 1;
	else dp[i][j] = 1;
      }
      else if(j == 0) dp[i][j] = 1;
      else {
	dp[i][j] = dp[i-1][j];
	if(j-i*i>=0) dp[i][j] += dp[i][j-i*i];
      }
    }
  }
  /*
  for(int i=0; i<=17; i++) {
    for(int j=0; j<=19; j++) printf("%3lld", dp[i][j]);
    cout << endl;
  }
  
  for(int i=0; i<=17; i++) {
    for(int j=20; j<=30; j++) printf("%3lld", dp[i][j]);
    cout << endl;
  }
  */

  int n;  
  
  while(cin>>n, n) {
    cout << dp[17][n] << endl;
  }
  
  return 0;
}
### Prompt

In CPP, your task is to solve 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;

long long dp[18][301];

int main() {
  
  for(int i=1; i<=17; i++) {
    for(int j=0; j<=300; j++) {
      if(i == 1) {
	if(j == 0) dp[1][0] = 1;
	else dp[i][j] = 1;
      }
      else if(j == 0) dp[i][j] = 1;
      else {
	dp[i][j] = dp[i-1][j];
	if(j-i*i>=0) dp[i][j] += dp[i][j-i*i];
      }
    }
  }
  /*
  for(int i=0; i<=17; i++) {
    for(int j=0; j<=19; j++) printf("%3lld", dp[i][j]);
    cout << endl;
  }
  
  for(int i=0; i<=17; i++) {
    for(int j=20; j<=30; j++) printf("%3lld", dp[i][j]);
    cout << endl;
  }
  */

  int n;  
  
  while(cin>>n, n) {
    cout << dp[17][n] << endl;
  }
  
  return 0;
}
```