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<set>
#include<algorithm>
using namespace std;
const int MAXN = 100000;
struct node{
	int num, key, vis;
};
node a[MAXN + 5];
bool operator < (node a, node b) {
	return a.key < b.key;
}
set<node>Set[MAXN + 5];
int fa[MAXN + 5];
int Find(int x) {
	if( fa[x] == x ) return x;
	return fa[x] = Find(fa[x]);
}
int main() {
	int N, M, cnt = 0;
	scanf("%d%d", &N, &M);
	for(int i=0;i<N;i++) {
		scanf("%d", &a[i].key);
		a[i].num = fa[i] = i;
	}
	int ned = ((N-1) - M) * 2;
	if( ned > N ) {
		printf("Impossible\n");
		return 0;
	}
	for(int i=1;i<=M;i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		fa[Find(u)] = Find(v);
	}
	for(int i=0;i<N;i++)
		Set[Find(i)].insert(a[i]);
	long long ans = 0;
	for(int i=0;i<N;i++) {
		if( Find(i) == i ) {
			if( cnt == ned ) continue;
			ans += Set[i].begin()->key;
			a[(*Set[i].begin()).num].vis = true;
			cnt++;
		}
	}
	sort(a, a+N);
	for(int i=0;i<N;i++) {
		if( cnt == ned ) break;
		if( a[i].vis ) continue;
		cnt++;
		ans += a[i].key;
	}
	printf("%lld\n", ans);
}