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<iostream>
#include<vector>
#include<algorithm>
using namespace std;

#define ran 101111

int n, m;
int a[ran];
vector<int> e[ran];
bool vis[ran];

vector<int> v, rem;

void dfs(int cur) {
	vis[cur] = true;
	v.push_back(a[cur]);
	for(int i=0;i<(int)e[cur].size();i++) {
		int nxt = e[cur][i];
		if (!vis[nxt]) dfs(nxt);
	}
}

int main() {
	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);
	}
	
	int todo = -2;
	long long ans = 0LL;
	rem.clear();
	for(int i=0;i<n;i++) {
		if (vis[i]) continue;
		v.clear();
		dfs(i);
		sort(v.begin(), v.end());
		ans += v[0];
		for(int j=1;j<(int)v.size();j++)
			rem.push_back(v[j]);
		todo++;
	}
	
	if (todo > (int)rem.size()) puts("Impossible");
	else if (todo<0) puts("0");
	else {
		sort(rem.begin(), rem.end());
		for(int i=0; i<todo; i++) ans += rem[i];
		cout<<ans<<endl;
	}

	return 0;
}