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 lint;
typedef pair<lint, int> pi;
typedef long long lint;
const int MAXN = 100005;

int n, m;
vector<int> cnum;
vector<int> gph[MAXN];
int vis[MAXN], a[MAXN];

void dfs(int x, int p){
	vis[x] = 1;
	cnum.push_back(a[x]);
	for(auto &i : gph[x]){
		if(!vis[i]) dfs(i, x);
	}
}

int main(){
	scanf("%d %d",&n,&m);
	if(m == n - 1){
		cout << 0 << endl;
		return 0;
	}
	for(int i=0; i<n; i++) scanf("%d",&a[i]);
	for(int i=0; i<m; i++){
		int s, e;
		scanf("%d %d",&s,&e);
		gph[s].push_back(e);
		gph[e].push_back(s);
	}
	int comp = 0;
	lint ans = 0;
	priority_queue<int, vector<int>, greater<int>> pq;
	for(int i=0; i<n; i++){
		if(vis[i]) continue;
		cnum.clear();
		dfs(i, -1);
		sort(cnum.begin(), cnum.end());
		ans += cnum[0];
		comp++;
		for(int i=1; i<cnum.size(); i++) pq.push(cnum[i]);
	}
	if((int)pq.size() < comp - 2){
		puts("Impossible");
		return 0;
	}
	for(int i=0; i<comp - 2; i++) ans += pq.top(), pq.pop();
	cout << ans << endl;
}
