#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<deque>
#include<map>
#include<set>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cctype>
using namespace std;
#define rep(i, j) for(int i = 0; i < j; i++)
#define f(i, k, j) for(int i = k; i < j; i++)
#define P pair<int, int>
const int INF = (1 << 26);
const int MAX_N = 301;

int dp[MAX_N], n;

void solve(){
    int c[16];
    for(int i = 2; i <= 17; i++) c[i - 2] = i * i;
    rep(i, MAX_N) dp[i] = 1;

    rep(i, 16){
        for(int j = c[i]; j < MAX_N; j++) dp[j] += dp[j - c[i]];
    }
}

int main(){
    solve();
    while(cin >>n && n != 0) cout <<dp[n] <<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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<deque>
#include<map>
#include<set>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cctype>
using namespace std;
#define rep(i, j) for(int i = 0; i < j; i++)
#define f(i, k, j) for(int i = k; i < j; i++)
#define P pair<int, int>
const int INF = (1 << 26);
const int MAX_N = 301;

int dp[MAX_N], n;

void solve(){
    int c[16];
    for(int i = 2; i <= 17; i++) c[i - 2] = i * i;
    rep(i, MAX_N) dp[i] = 1;

    rep(i, 16){
        for(int j = c[i]; j < MAX_N; j++) dp[j] += dp[j - c[i]];
    }
}

int main(){
    solve();
    while(cin >>n && n != 0) cout <<dp[n] <<endl;
    return 0;
}
```