CF708B Recover the String 构造

For each string s consisting of characters '0' and '1' one can define four integers a₀₀, a₀₁, a₁₀ and a₁₁, where a_xy is the number of subsequences of length 2 of the string s equal to the sequence {x, y}.

In these problem you are given four integers a₀₀, a₀₁, a₁₀, a₁₁ and have to find any non-empty string s that matches them, or determine that there is no such string. One can prove that if at least one answer exists, there exists an answer of length no more than 1 000 000.

Input

The only line of the input contains four non-negative integers a₀₀, a₀₁, a₁₀ and a₁₁. Each of them doesn't exceed 10⁹.

Output

If there exists a non-empty string that matches four integers from the input, print it in the only line of the output. Otherwise, print "Impossible". The length of your answer must not exceed 1 000 000.

Examples

Input

Copy

1 2 3 4

Output

Copy

Impossible

Input

Copy

1 2 2 1

Output

Copy

0110

 

 

首先从a00和a11可以求出0和1的数量；

假设目前为0000...001111..；

显然a01=num0*num1，a10=0;

那么我们移动一个1去左边，a10++,a01--;但总数还是不变；

那么合理的解必须是a10+a01=num0*num1;

值得注意的是：当00或11=0时，0或1可能有1个也可能没有；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/
int a00, a01, a10, a11;

int main()
{
	//ios::sync_with_stdio(0);
	rdint(a00); rdint(a01); rdint(a10); rdint(a11);
	int k = 1, r = 1;
	while (k*(k - 1) / 2 < a00)k++;
	while (r*(r - 1) / 2 < a11)r++;
//	cout << k << ' ' << r << endl;
	int fg = 1;
	if (k*(k - 1) / 2 != a00 || r * (r - 1) / 2 != a11) {
		cout << "Impossible" << endl; return 0;
	}
	
	else if (a00 == 0 && a11 == 0) {
		if (a01 == 0 && a10 == 0)cout << 0 << endl;
		else if (a01 == 0 && a10 == 1)cout << "10" << endl;
		else if (a01 == 1 && a10 == 0)cout << "01" << endl;
		else cout << "Impossible" << endl;
		return 0;
	}
	else if (a00 == 0) {
		if (a01 == 0 && a10 == 0) {
			while (r--)cout << '1';

		}
		else if (a01 + a10 == r) {
			while (a10--)cout << '1';
			cout << 0;
			while (a01--)cout << '1';
		}
		else fg = 0;
	}
	else if (a11 == 0) {
		if (a10 == 0 && a01 == 0) {
			while (k--)cout << '0';
		}
		else if (a10 + a01 == k) {
			while (a01--)cout << 0;
			cout << 1;
			while (a10--)cout << 0;
		}
		else fg = 0;
	}

	else {
		if (a01 + a10 == k * r) {
			while (a01) {
				while (r > a01) {
					cout << 1;
					r--;
				}
				cout << 0;
				k--; a01 -= r;
			}
			while (r--)cout << 1;
			while (k--)cout << 0;
		}
		else fg = 0;
	}
	if (fg == 0)cout << "Impossible" << endl;
	return 0;
}