#include <iostream>
#include <cstdio>
#include <algorithm>
#include <array>
#include <vector>

using namespace std;

int main() {

  array<array<int, 18>, 301> dp;
  for ( auto ar : dp ) {
	fill( ar.begin(), ar.end(), 0 );
  }
  // dp[ 現在の金 ][ 最大sq ]
  dp[ 0 ][ 1 ] = 1;
  for ( int i = 1; i <= 300; i++ ){
	for ( int j = 1; j <= 17; j++ ) {
	  for ( int k = 0; k <= j; k++ ) {
		if ( i - j * j >= 0 ) {
		  dp[ i ][ j ] += dp[ i - j * j ][ k ];
		}
	  }
	}	  
  }

  array<int, 301> ans;
  fill( ans.begin(), ans.end(), 0 );
  for ( int i = 1; i <= 300; i++ ) {
	for ( int j = 1; j <= 17; j++ ) {
	  ans[ i ] += dp[ i ][ j ];
	}
  }
  
  while( true ) {
	int n = 0;
	cin >> n;
	if ( n == 0 ) break;
	cout << ans[ n ] << endl;
  }
}
  
### Prompt

Generate 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>
#include <cstdio>
#include <algorithm>
#include <array>
#include <vector>

using namespace std;

int main() {

  array<array<int, 18>, 301> dp;
  for ( auto ar : dp ) {
	fill( ar.begin(), ar.end(), 0 );
  }
  // dp[ 現在の金 ][ 最大sq ]
  dp[ 0 ][ 1 ] = 1;
  for ( int i = 1; i <= 300; i++ ){
	for ( int j = 1; j <= 17; j++ ) {
	  for ( int k = 0; k <= j; k++ ) {
		if ( i - j * j >= 0 ) {
		  dp[ i ][ j ] += dp[ i - j * j ][ k ];
		}
	  }
	}	  
  }

  array<int, 301> ans;
  fill( ans.begin(), ans.end(), 0 );
  for ( int i = 1; i <= 300; i++ ) {
	for ( int j = 1; j <= 17; j++ ) {
	  ans[ i ] += dp[ i ][ j ];
	}
  }
  
  while( true ) {
	int n = 0;
	cin >> n;
	if ( n == 0 ) break;
	cout << ans[ n ] << endl;
  }
}
  
```