#include <iostream>
#include <memory>
#include <memory.h>
#include <cmath>
#include <fstream>
#include <cmath>
#include <numeric>
#include <vector>
#include <stack>
#include <string>
#include <queue>
#include <sstream>
#include <cstdlib>
#include <cassert>
#include <cstdio>
#include <map>
#include <iomanip>
#include <list>
#include <cctype>
#include <algorithm>
#include <complex>
#define USE_MATH_DEFINES
using namespace std;

typedef complex<double> xy_t;
typedef pair<xy_t, xy_t> line;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, P> PP;
typedef pair<int, PP> PPP;
const int INF = 1 << 29;
const double EPS = 1E-10;

namespace std{
	bool operator < (const xy_t &a, const xy_t &b){
		return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
	}
}

double cross(const xy_t &a, const xy_t &b){
	return imag(conj(a) * b);
}

double dot(const xy_t &a, const xy_t &b){
	return real(conj(a)*b);
}

xy_t projection(const line &l, const xy_t &p){
	double t = dot(l.second - l.first, p - l.first) / norm(l.second - l.first);
	return l.first + t * (l.second - l.first);
}

int ccw(xy_t a, xy_t b, xy_t c){
	b -= a; c -= a;
	if(cross(b, c) > 0) return 1;
	if(cross(b, c) < 0) return -1;
	if(dot(b, c) < 0) return +2;
	if(norm(b) < norm(c) ) return -2;
	return 0;
}

bool intersectSP(const line &s, const xy_t &p){
	return abs(s.first - p) + abs(s.second - p) - abs(s.second - s.first) < EPS;
}

bool intersectSS(const line &s, const line &t){
	return ccw(s.first, s.second, t.first) * ccw(s.first, s.second, t.second) <= 0 &&
		ccw(t.first, t.second, s.first) * ccw(t.first, t.second, s.second) <= 0;
}


double distanceSP(const line &s, const xy_t p){
	xy_t r = projection(s, p);
	if(intersectSP(s, r)) return abs(r - p);
	return min(abs(s.first - p), abs(s.second - p));
}

double distanceSS(const line &s, const line &t){
	if(intersectSS(s, t)) return 0.0;
	return min(min(distanceSP(s, t.first), distanceSP(s, t.second)),
		min(distanceSP(t, s.first), distanceSP(t, s.second)));
}

xy_t crosspoint(const line &l, const line &m){
	double a = cross(l.second - l.first, m.second - m.first);
	double b = cross(l.second - l.first, l.second - m.first);
	if(abs(a) < EPS && abs(b) < EPS) return m.first;
	if(abs(a) < EPS) assert(false);
	return m.first + b / a * (m.second - m.first);
}

int main(){
	int dp[20][1000];
	int n;
	while(cin >> n && n){
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1;
		for(int i = 1; i <= 17; i++){
			for(int j = 0; j <= n; j++){
				dp[i][j] = dp[i-1][j];
				if(j >= i * i){
					dp[i][j] += dp[i][j-i*i];
				}
			}
		}
		cout << dp[17][n] << endl;
	}
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 <memory>
#include <memory.h>
#include <cmath>
#include <fstream>
#include <cmath>
#include <numeric>
#include <vector>
#include <stack>
#include <string>
#include <queue>
#include <sstream>
#include <cstdlib>
#include <cassert>
#include <cstdio>
#include <map>
#include <iomanip>
#include <list>
#include <cctype>
#include <algorithm>
#include <complex>
#define USE_MATH_DEFINES
using namespace std;

typedef complex<double> xy_t;
typedef pair<xy_t, xy_t> line;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, P> PP;
typedef pair<int, PP> PPP;
const int INF = 1 << 29;
const double EPS = 1E-10;

namespace std{
	bool operator < (const xy_t &a, const xy_t &b){
		return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
	}
}

double cross(const xy_t &a, const xy_t &b){
	return imag(conj(a) * b);
}

double dot(const xy_t &a, const xy_t &b){
	return real(conj(a)*b);
}

xy_t projection(const line &l, const xy_t &p){
	double t = dot(l.second - l.first, p - l.first) / norm(l.second - l.first);
	return l.first + t * (l.second - l.first);
}

int ccw(xy_t a, xy_t b, xy_t c){
	b -= a; c -= a;
	if(cross(b, c) > 0) return 1;
	if(cross(b, c) < 0) return -1;
	if(dot(b, c) < 0) return +2;
	if(norm(b) < norm(c) ) return -2;
	return 0;
}

bool intersectSP(const line &s, const xy_t &p){
	return abs(s.first - p) + abs(s.second - p) - abs(s.second - s.first) < EPS;
}

bool intersectSS(const line &s, const line &t){
	return ccw(s.first, s.second, t.first) * ccw(s.first, s.second, t.second) <= 0 &&
		ccw(t.first, t.second, s.first) * ccw(t.first, t.second, s.second) <= 0;
}


double distanceSP(const line &s, const xy_t p){
	xy_t r = projection(s, p);
	if(intersectSP(s, r)) return abs(r - p);
	return min(abs(s.first - p), abs(s.second - p));
}

double distanceSS(const line &s, const line &t){
	if(intersectSS(s, t)) return 0.0;
	return min(min(distanceSP(s, t.first), distanceSP(s, t.second)),
		min(distanceSP(t, s.first), distanceSP(t, s.second)));
}

xy_t crosspoint(const line &l, const line &m){
	double a = cross(l.second - l.first, m.second - m.first);
	double b = cross(l.second - l.first, l.second - m.first);
	if(abs(a) < EPS && abs(b) < EPS) return m.first;
	if(abs(a) < EPS) assert(false);
	return m.first + b / a * (m.second - m.first);
}

int main(){
	int dp[20][1000];
	int n;
	while(cin >> n && n){
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1;
		for(int i = 1; i <= 17; i++){
			for(int j = 0; j <= n; j++){
				dp[i][j] = dp[i-1][j];
				if(j >= i * i){
					dp[i][j] += dp[i][j-i*i];
				}
			}
		}
		cout << dp[17][n] << endl;
	}
	return 0;
}
```