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 <algorithm>
#include <queue>
using namespace std;

typedef long long ll;
typedef pair<ll, int> Node;

ll cost[100005], ans;
int p[100005];

priority_queue<Node, vector<Node>, greater<Node>> st[100005], q;

int find(int a) {
	if (a == p[a]) return a;
	return p[a] = find(p[a]);
}

bool merge(int a, int b) {
	a = find(a), b = find(b);
	if (a == b) return 0;
	if (st[a].size() < st[b].size()) swap(a, b);
	while (!st[b].empty()) {
		st[a].push(st[b].top()); st[b].pop();
	}
	p[b] = a;
	return 1;
}

int main() {
	int N, M; scanf("%d %d", &N, &M);
	for (int i = 0; i < N; i++) {
		scanf("%lld", cost + i);
		p[i] = i;
		st[i].push({cost[i], i});
	}
	int remain = N - 2 - M;
	if (remain < 0) return !printf("0");
	while (M--) {
		int a, b; scanf("%d %d", &a, &b);
		if (!merge(a, b)) return !printf("Impossible");
	}

	for (int i = 0; i < N; i++) if (i == find(i)) {
		ans += st[i].top().first; st[i].pop();
		while (!st[i].empty()) {
			q.push(st[i].top());
			st[i].pop();
		}
	}
	while (remain && !q.empty()) {
		ans += q.top().first; q.pop();
		remain--;
	}
	if (remain) puts("Impossible");
	else printf("%lld", ans);
}