#include "bits/stdc++.h"
/*
#include <iostream>
#include <vector>
#include <map>
#include <utility>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <numeric>
#include <complex>
#include <bitset>
#include <functional>
#include <stack>
#include <regex>
#include <tuple>
*/
#define int long long
#define REP(i,a,n) for(int i=a;i<n;++i)
#define rep(i,n) REP(i,0,n)
#define REV(i,a,n) for(int i=n;i>=a;--i)
#define all(e) e.begin(),e.end()
#define rall(e) e.rbegin(),e.rend()
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define mod 1000000007
#define show(n) cerr<<#n<<" = "<<n<<endl
#define showp(n) cerr<<n.fs<<", "<<n.sc<<endl
#define shows(n) for(auto z:n){cerr<<z<<", ";}cerr<<endl
#define showsp(n) for(auto z:n){cerr<<z.fs<<" "<<z.sc<<", "}cerr<<endl

#define yes printf("Yes\n")
#define no printf("No\n")
#define case(i) printf("Case #%lld: ",i)

using namespace std;

using vi=vector<int>;
using pint=pair<int,int>;

const int INF=1LL<<55;

int dp[18][301];
int n;

int rec(int idx, int mul){
    if(mul<0) return 0;
    if(idx>17) return mul==0;
    if(dp[idx][mul]) return dp[idx][mul];
    return dp[idx][mul]=rec(idx + 1, mul) + rec(idx, mul-idx*idx);
}

void solve(){
    cout<<rec(1,n)<<endl;
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    while(cin>>n,n){
        solve();
    }

    return 0;
}

### 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"
/*
#include <iostream>
#include <vector>
#include <map>
#include <utility>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <numeric>
#include <complex>
#include <bitset>
#include <functional>
#include <stack>
#include <regex>
#include <tuple>
*/
#define int long long
#define REP(i,a,n) for(int i=a;i<n;++i)
#define rep(i,n) REP(i,0,n)
#define REV(i,a,n) for(int i=n;i>=a;--i)
#define all(e) e.begin(),e.end()
#define rall(e) e.rbegin(),e.rend()
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define mod 1000000007
#define show(n) cerr<<#n<<" = "<<n<<endl
#define showp(n) cerr<<n.fs<<", "<<n.sc<<endl
#define shows(n) for(auto z:n){cerr<<z<<", ";}cerr<<endl
#define showsp(n) for(auto z:n){cerr<<z.fs<<" "<<z.sc<<", "}cerr<<endl

#define yes printf("Yes\n")
#define no printf("No\n")
#define case(i) printf("Case #%lld: ",i)

using namespace std;

using vi=vector<int>;
using pint=pair<int,int>;

const int INF=1LL<<55;

int dp[18][301];
int n;

int rec(int idx, int mul){
    if(mul<0) return 0;
    if(idx>17) return mul==0;
    if(dp[idx][mul]) return dp[idx][mul];
    return dp[idx][mul]=rec(idx + 1, mul) + rec(idx, mul-idx*idx);
}

void solve(){
    cout<<rec(1,n)<<endl;
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    while(cin>>n,n){
        solve();
    }

    return 0;
}

```