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<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int N=100005;
typedef long long ll;

int n,fa[N],a[N],m,id[N];
int find(int x) {return x==fa[x]?x:fa[x]=find(fa[x]);}
bool b[N];
ll ans;

int tmp[N];
bool cmp(int x,int y) {return a[x]<a[y];}

int main()
{
	scanf("%d%d",&n,&m);
	if (m==n-1) {puts("0");return 0;}
	for (int i=1;i<=n;i++) scanf("%d",&a[i]),fa[i]=i;
	for (int u,v,i=1;i<=m;i++)
	{
		scanf("%d%d",&u,&v);u++,v++;
		fa[find(u)]=find(v);
	}
	
	int cnt=0;
	for (int i=1;i<=n;i++) if (fa[i]==i) id[i]=i;
	for (int i=1;i<=n;i++) if (a[i]<a[id[find(i)]]) id[fa[i]]=i;
	for (int i=1;i<=n;i++) 
	{
		if (fa[i]==i) b[id[i]]=true,ans+=a[id[i]],cnt++;
		tmp[i]=i;
	}

	sort(tmp+1,tmp+n+1,cmp);
	for (int i=1;i<=n&&cnt<2*(n-1-m);i++) 
	if (!b[tmp[i]]) ans+=a[tmp[i]],++cnt;
	
	if (cnt<2*(n-1-m)) puts("Impossible");
	else printf("%lld\n",ans);
	return 0;
}
