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 <bits/stdc++.h>

using namespace std;

typedef long long ll;

ll P[101010], A[101010];
vector <ll> S[101010], V, K;
ll n, m, ans;

ll find(ll p) { return p == P[p]? p : P[p] = find(P[p]); }

int main()
{
	ll i, a, b, k;
	
	scanf("%lld%lld", &n, &m);
	
	for(i=0; i<n; i++){
		scanf("%lld", A+i);
		P[i] = i;
	}
	
	for(i=1; i<=m; i++){
		scanf("%lld%lld", &a, &b);
		P[find(a)] = find(b);
	}
	
	for(i=0; i<n; i++){
		S[find(i)].push_back(A[i]);
		if(i == P[i]) K.push_back(i);
	}
	
	if((K.size() - 1) * 2 > n){
		printf("Impossible\n");
		return 0;
	}
	else if(K.size() == 1){
		printf("0\n");
		return 0;
	}
	
	k = K.size() - 2;
	
	for(int t: K){
		if(S[t].size()){
			sort(S[t].begin(), S[t].end());
			ans += S[t][0];
			for(i=1; i<S[t].size(); i++) V.push_back(S[t][i]);
		}
	}
	
	if(V.size()) sort(V.begin(), V.end());
	
	for(i=0; i<k; i++) ans += V[i];
	
	printf("%lld\n", ans);
	
	return 0;
}