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;
typedef pair <int, int> pii;

const int MAXN = 1e5 + 5;

int N, M;
int a[MAXN];
pii p[MAXN];
vector <int> E[MAXN];
bool bio[MAXN], used[MAXN];
int opt, cnt;
ll sol;

void load() {
	scanf("%d%d", &N, &M);
	for (int i = 0; i < N; i++)	
		scanf("%d", a + i);
	for (int i = 0; i < M; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		E[x].push_back(y);
		E[y].push_back(x);
	}
}

void dfs(int x) {
	bio[x] = true;
	if (a[x] < a[opt])
		opt = x;
	for (auto it : E[x])
		if (!bio[it])
			dfs(it);
}

void add(int x) {
	sol += a[x];
	used[x] = true;
	cnt++;
}

void solve() {
	a[N] = 2e9;
	int comp = -1;
	for (int i = 0; i < N; i++) {
		if (bio[i])
			continue;
		opt = N;
		dfs(i);
		add(opt);
		comp++;
	}
	
	if (!comp) {
		puts("0");
		return;
	}
	
	for (int i = 0; i < N; i++)
		p[i] = {a[i], i};
		
	sort(p, p + N);
	
	for (int i = 0; i < N && cnt < 2 * comp; i++)
		if (!used[p[i].second])
			add(p[i].second);
	
	if (cnt < 2 * comp)
		puts("Impossible");
	else
		printf("%lld\n", sol);
}

int main() {
	load();
	solve();
	return 0;
}