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 <vector>
#include <algorithm>

using namespace std;

const int NMAX = 1e5;

int n,m;
vector<int> graph[NMAX + 5];
int val[NMAX + 5];
bool viz[NMAX + 5];

int dfs(int nod){
	viz[nod] = 1;
	int ans = nod;
	for(auto it:graph[nod]){
		if(!viz[it]){
			int tmp = dfs(it);
			if(val[tmp] < val[ans]){
				ans = tmp;
			}
		}
	}
	return ans;
}

int main(){
	
	scanf("%d %d",&n,&m);
	
	for(int i = 0;i < n;i++){
		scanf("%d",&val[i]);
	}
	
	for(int i = 0;i < m;i++){
		int x,y;
		scanf("%d %d",&x,&y);
		graph[x].push_back(y);
		graph[y].push_back(x);
	}
	
	if(2 * (n - m - 1) > n){
		printf("Impossible");
		return 0;
	}
	
	if(m == n - 1){
		printf("0");
		return 0;
	}
	
	long long ans = 0;
	
	for(int i = 0;i < n;i++){
		if(!viz[i]){
			int tmp = dfs(i);
			ans += val[tmp];
			val[tmp] = 1 << 30;
		}
	}
	
	sort(val,val + n);
	
	for(int i = 0;i < n - m - 2;i++){
		ans += val[i];
	}
	
	printf("%lld",ans);

	return 0;
}