#include<iostream>
#include<algorithm>

using namespace std;

int dp[18][301];
int value[18];


void makedata(void){
  for(int i = 0; i < 18; i++){
    value[i] = i*i;
  }
}

void makedp(void){

  for(int i = 0; i < 18; i++){
    for(int j = 0; j < 300; j++){
      dp[i][j] = 0;
    }
  }

  dp[0][0] = 1;

  for(int i = 1; i < 18; i++){
    for(int j = 0; j < 301; j++){
      dp[i][j] += dp[i-1][j];
      if(j - value[i] >= 0) dp[i][j] += dp[i][j-value[i]];
    }
  }
}

int main(){
  makedata();
  makedp();
  int in;
  while(cin >> in){
    if(in == 0) break;
    cout << dp[17][in]<<endl;
  }
  return 0;
}
### Prompt

Develop a solution in cpp 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>
#include<algorithm>

using namespace std;

int dp[18][301];
int value[18];


void makedata(void){
  for(int i = 0; i < 18; i++){
    value[i] = i*i;
  }
}

void makedp(void){

  for(int i = 0; i < 18; i++){
    for(int j = 0; j < 300; j++){
      dp[i][j] = 0;
    }
  }

  dp[0][0] = 1;

  for(int i = 1; i < 18; i++){
    for(int j = 0; j < 301; j++){
      dp[i][j] += dp[i-1][j];
      if(j - value[i] >= 0) dp[i][j] += dp[i][j-value[i]];
    }
  }
}

int main(){
  makedata();
  makedp();
  int in;
  while(cin >> in){
    if(in == 0) break;
    cout << dp[17][in]<<endl;
  }
  return 0;
}
```