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;

#define N 120000
#define INF 0x3f3f3f3f

int n,m,x,y,xb,cnt,pos[N],c[N],a[N],fa[N],f[N];
long long ans;
multiset<int> g[N];

bool cmp(int x,int y){return f[x]>f[y];}

int gfa(int x){return x==fa[x]?x:fa[x]=gfa(fa[x]);}

int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int i=1;i<=m;i++){
		scanf("%d%d",&x,&y); x++; y++;
		x=gfa(x); y=gfa(y);
		fa[x]=y;
	}
	for (int i=1;i<=n;i++) f[gfa(i)]++,c[i]=i;
	sort(c+1,c+n+1,cmp);
	for (int i=1;i<=n;i++){
		if (f[c[i]]==0) break;
		pos[c[i]]=++cnt;
	}
	for (int i=1;i<=n;i++) g[pos[gfa(i)]].insert(a[i]);
	for (int i=2;i<=cnt;i++){
		if (g[1].empty()){puts("Impossible"); return 0;}
		ans+=*g[1].begin()+*g[i].begin();
		g[1].erase(g[1].begin()); g[i].erase(g[i].begin());
		for (set<int>::iterator j=g[i].begin();j!=g[i].end();j++)
			g[1].insert(*j);
		g[i].clear();
	}
	printf("%lld\n",ans);
	
	return 0;
}
			