#include<bits/stdc++.h>
using namespace std;

int main(){

    while(1){

        int n; cin >> n;
        if(!n) break;
        vector<int> a(17);
        for(int i = 1; i <= 17; i++){
            a[i - 1] = i * i;
        }


        vector<vector<int> > dp(18, vector<int> (1000, 0));
        dp[0][0] = 1;
        for(int i = 0; i < 17; i++){
            for(int j = 0; j < 500; j++){
                for(int k = 0;;k++){

                    int next_n = j + k * a[i];
                    if(next_n > n) break;

                    dp[i + 1][next_n] += dp[i][j];

                }
            }
        }

        cout << dp[17][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<bits/stdc++.h>
using namespace std;

int main(){

    while(1){

        int n; cin >> n;
        if(!n) break;
        vector<int> a(17);
        for(int i = 1; i <= 17; i++){
            a[i - 1] = i * i;
        }


        vector<vector<int> > dp(18, vector<int> (1000, 0));
        dp[0][0] = 1;
        for(int i = 0; i < 17; i++){
            for(int j = 0; j < 500; j++){
                for(int k = 0;;k++){

                    int next_n = j + k * a[i];
                    if(next_n > n) break;

                    dp[i + 1][next_n] += dp[i][j];

                }
            }
        }

        cout << dp[17][n] << endl;
    }
}

```