#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;

int main(){
  int n;
  int coins[17];
  for(int i = 0; i < 17; i++)
    coins[i]=(i+1)*(i+1);
  int dp[2][400];
  //fill(dp,dp+400,0);
  memset(dp,0,sizeof(dp));
  dp[0][0]=1;
  // ツづづ個硬ツ嘉敖づーツ使ツつ、ツつゥ
  for(int j = 0; j < 17; j++){
    memset(dp[(j+1)%2],0,sizeof(dp[(j+1)%2]));
    for(int i = 0; i <= 300; i++){
      for(int k = 0; k <= 400; k++){
	if(i-k*coins[j]>=0)
	  dp[(j+1)%2][i]+=dp[j%2][i-k*coins[j]];
	else
	  break;
      }
    }
  }
  while(cin>>n&&n!=0)
    cout<<dp[1][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>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;

int main(){
  int n;
  int coins[17];
  for(int i = 0; i < 17; i++)
    coins[i]=(i+1)*(i+1);
  int dp[2][400];
  //fill(dp,dp+400,0);
  memset(dp,0,sizeof(dp));
  dp[0][0]=1;
  // ツづづ個硬ツ嘉敖づーツ使ツつ、ツつゥ
  for(int j = 0; j < 17; j++){
    memset(dp[(j+1)%2],0,sizeof(dp[(j+1)%2]));
    for(int i = 0; i <= 300; i++){
      for(int k = 0; k <= 400; k++){
	if(i-k*coins[j]>=0)
	  dp[(j+1)%2][i]+=dp[j%2][i-k*coins[j]];
	else
	  break;
      }
    }
  }
  while(cin>>n&&n!=0)
    cout<<dp[1][n]<<endl;
  return 0;
}
```