You are given a forest with N vertices and M edges. The vertices are numbered 0 through N-1. The edges are given in the format (x_i,y_i), which means that Vertex x_i and y_i are connected by an edge.

Each vertex i has a value a_i. You want to add edges in the given forest so that the forest becomes connected. To add an edge, you choose two different vertices i and j, then span an edge between i and j. This operation costs a_i + a_j dollars, and afterward neither Vertex i nor j can be selected again.

Find the minimum total cost required to make the forest connected, or print `Impossible` if it is impossible.

Constraints

* 1 ≤ N ≤ 100,000
* 0 ≤ M ≤ N-1
* 1 ≤ a_i ≤ 10^9
* 0 ≤ x_i,y_i ≤ N-1
* The given graph is a forest.
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N M
a_0 a_1 .. a_{N-1}
x_1 y_1
x_2 y_2
:
x_M y_M


Output

Print the minimum total cost required to make the forest connected, or print `Impossible` if it is impossible.

Examples

Input

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6


Output

7


Input

5 0
3 1 4 1 5


Output

Impossible


Input

1 0
5


Output

0
#include <cstdio>
#include <cstdlib>
#include <algorithm>

const int maxn = 1e5 + 20;

int n, m, A[maxn], F[maxn];
std::pair<int,int> M[maxn];
bool used[maxn];

int find(int x)
{
	return F[x] == x ? x : (F[x] = find(F[x]));
}
template<class TAT>void checkmin(TAT &x,const TAT &y)
{
	if(y < x) x = y;
}
int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 0; i < n; i++) scanf("%d", &A[i]);
	
	for(int i = 0; i < n; i++) F[i] = i;
	
	for(int i = 0, x, y; i < m; i++)
	{
		scanf("%d%d", &x, &y);
		F[find(x)] = find(y);
	}
	
	for(int i = 0; i < n; i++)
		M[i] = std::make_pair(A[i], i);
	
	for(int i = 0; i < n; i++)
		checkmin(M[find(i)], M[i]);
	
	long long ans = 0;
	int cnt = 0;
	
	for(int i = 0; i < n; i++)
		if(find(i) == i)
		{
			ans += M[i].first, cnt++;
			used[M[i].second] = true;
		}

	if(cnt == 1)
	{
		printf("0\n");
	}
	else if(n < (cnt - 1) * 2)
	{
		printf("Impossible\n");
	}
	else
	{
		for(int i = 0; i < n; i++)
			M[i] = std::make_pair(A[i], i);
		std::sort(M, M + n);
		
		int rem = (cnt - 1) * 2 - cnt;
		
		for(int i = 0; i < n && rem; i++)
		{
			if(used[M[i].second]) continue;
			ans += M[i].first, rem--;
		}
		
		printf("%lld\n", ans);
	}
	
	return 0;
}