#include <iostream>
#include <cstring>

using namespace std;
int dp[18][301][301];
int hoge;
int main(){
    memset(dp,0,sizeof(dp));
    for(int i=0;i<301;++i) dp[0][i][i] = 1;
    
    for(int i=1;i<18;i++){
        for(int j=0;j<301;j++){
            for(int k=0;k<301;++k){
                int cnt=0;
                while((i+1)*(i+1)*cnt+k<301){
                    dp[i][j+cnt][(i+1)*(i+1)*cnt+k] += dp[i-1][j][k];
                    cnt++;
                }
            }
        }
    }

    int n;
    while(cin >> n && n){
        int ret=0;
        for(int i=0;i<301;++i) ret+= dp[17][i][n];
        cout << ret << endl;
    }
}
### Prompt

Please create a solution in cpp 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 <cstring>

using namespace std;
int dp[18][301][301];
int hoge;
int main(){
    memset(dp,0,sizeof(dp));
    for(int i=0;i<301;++i) dp[0][i][i] = 1;
    
    for(int i=1;i<18;i++){
        for(int j=0;j<301;j++){
            for(int k=0;k<301;++k){
                int cnt=0;
                while((i+1)*(i+1)*cnt+k<301){
                    dp[i][j+cnt][(i+1)*(i+1)*cnt+k] += dp[i-1][j][k];
                    cnt++;
                }
            }
        }
    }

    int n;
    while(cin >> n && n){
        int ret=0;
        for(int i=0;i<301;++i) ret+= dp[17][i][n];
        cout << ret << endl;
    }
}
```