#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 18;
const int MAX_N = 301;

int value[N];
int dp[N][MAX_N];

void make_Table() {
  for(int i = 1 ; i < N ; i++) {
    value[i] = i * i;
  }
}

void solve() {
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  
  for(int i = 1 ; i < N ; i++) {
    for(int j = 0 ; j < MAX_N ; j++) {
      int prevalue = j - value[i];
      
      dp[i][j] = dp[i-1][j];
      if(prevalue >= 0) dp[i][j] += dp[i][prevalue];
    }
  }
}

int main() {
  int n;
  make_Table();
  solve();
  while(cin >> n, n) {
    cout << dp[N-1][n] << endl;
  }
  return 0;
}
### 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 <cstring>
using namespace std;

const int N = 18;
const int MAX_N = 301;

int value[N];
int dp[N][MAX_N];

void make_Table() {
  for(int i = 1 ; i < N ; i++) {
    value[i] = i * i;
  }
}

void solve() {
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  
  for(int i = 1 ; i < N ; i++) {
    for(int j = 0 ; j < MAX_N ; j++) {
      int prevalue = j - value[i];
      
      dp[i][j] = dp[i-1][j];
      if(prevalue >= 0) dp[i][j] += dp[i][prevalue];
    }
  }
}

int main() {
  int n;
  make_Table();
  solve();
  while(cin >> n, n) {
    cout << dp[N-1][n] << endl;
  }
  return 0;
}
```