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<iostream>
#include<cstdio>
#include<vector>
#include<set>

using namespace std;

const int MAXN = 100005;

int n, m, cnt;
long long sol;
int l[MAXN], bio[MAXN], mn[MAXN];
vector <int> v[MAXN];
multiset <int> s;

void dfs (int x, int boja) {
	if (bio[x]) return;
	bio[x]=boja;
	if (mn[boja]==0 || l[x]<l[mn[boja]]) mn[boja]=x;
	for (int i=0; i<v[x].size(); i++) {
		dfs(v[x] [i], boja);
	}
}

int main () {
	cin >> n >> m;
	for (int i=1; i<=n; i++) {
		scanf("%d", &l[i]);
	}
	for (int i=0; i<m; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		a++; b++;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for (int i=1; i<=n; i++) {
		if (!bio[i]) {
			cnt++;
			dfs(i, cnt);
		}
	}
	if (cnt==1) {
		cout << 0;
		return 0;
	}
	for (int i=1; i<=n; i++) {
		int boja=bio[i];
		if (i==mn[boja]) sol+=l[i]; else s.insert(l[i]);
	}
	if (s.size()<cnt-2) {
		cout << "Impossible";
		return 0;
	}
	int br=0;
	for (set <int> :: iterator it=s.begin(); it!=s.end(); it++) {
		if (br==cnt-2) break;
		sol+=*it;
		br++;
	}
	cout << sol;
	return 0;
}


