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;

vector<int> q,sq,e[100010];
int a[100010];
bool v[100010];

void dfs(int x) {
	v[x] = true;
	if (q.empty() || a[x] >= q[0]) q.push_back(a[x]);
	else {
		q.push_back(q[0]);
		q[0] = a[x];
	}
	for (int i = 0; i < e[x].size(); i++) 
		if (!v[e[x][i]]) dfs(e[x][i]);
}

int main() {
	int n,m;
	scanf("%d%d",&n,&m);
	for (int i = 0; i < n; i++) scanf("%d",&a[i]);
	
	while (m--) {
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	
	long long ans = 0, tot = 0;
	for (int i = 0; i < n; i++) if (!v[i]) {
		tot++;
		dfs(i);
		ans += q[0];
		for (int i = 1; i < q.size(); i++) sq.push_back(q[i]);
		q.clear();
	}
	
	if (tot < 2) {
		printf("0\n");
		return 0;	
	}
	
	if (sq.size() < tot - 2) {
		printf("Impossible\n");
		return 0;
	}
	
	sort(sq.begin(),sq.end());
	for (int i = 0; i < tot - 2; i++) ans += sq[i];
	printf("%lld\n",ans);
}