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

const int s[]={0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289};


int f(int n,int m){
    if(n==0)return 1;
    int ans=1;
    vector<int> t(18,0);
    t[1]=n;
    for (int i = 2; i < m; ++i) {
        if(n<s[i])break;
        while(n-s[i]*t[i]>=s[i]){
            ++t[i];
            ans+=f(n-s[i]*t[i],i);
        }
    }
    return ans;
}

int main(){
    int n;

    while(true){
        cin>>n;
        if(n==0)break;
        int ans=1;
        ans=f(n,18);
        cout<<ans<<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 <bits/stdc++.h>
using namespace std;

const int s[]={0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289};


int f(int n,int m){
    if(n==0)return 1;
    int ans=1;
    vector<int> t(18,0);
    t[1]=n;
    for (int i = 2; i < m; ++i) {
        if(n<s[i])break;
        while(n-s[i]*t[i]>=s[i]){
            ++t[i];
            ans+=f(n-s[i]*t[i],i);
        }
    }
    return ans;
}

int main(){
    int n;

    while(true){
        cin>>n;
        if(n==0)break;
        int ans=1;
        ans=f(n,18);
        cout<<ans<<endl;
    }
    return 0;
}

```