#include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define repp(i, n) rep(i, 0, n)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
using namespace std;

int dp[310];

signed main(){
    dp[0] = 1;
    rep(i, 1, 18){
        rep(j, 0, 310){
            if(j - i*i >= 0){
                dp[j] += dp[j - i*i];
            }
        }
    }

    int n;
    while(cin >> n, n){
        cout << dp[n] << endl;
    }
}
### Prompt

Create a solution in CPP for 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>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define repp(i, n) rep(i, 0, n)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
using namespace std;

int dp[310];

signed main(){
    dp[0] = 1;
    rep(i, 1, 18){
        rep(j, 0, 310){
            if(j - i*i >= 0){
                dp[j] += dp[j - i*i];
            }
        }
    }

    int n;
    while(cin >> n, n){
        cout << dp[n] << endl;
    }
}
```