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<algorithm>
#include<cstdio>
using namespace std;
int t[101000];
int val[101000],b[101000],c[101000],k2=0;
int h[101000],nxt[201000],K=0,to[201000];
void ins(int u,int v){nxt[++K]=h[u];h[u]=K;to[K]=v;}
int Min;
void dfs(int x,int o)
{
	t[x]=o;Min=min(Min,val[x]);c[++k2]=x;
	for(int i=h[x];i;i=nxt[i])if(!t[to[i]])dfs(to[i],o);
}
int main()
{
	int n,m;scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)scanf("%d",&val[i]);
	for(int i=1;i<=m;i++)
	{
		int u,v;scanf("%d%d",&u,&v);u++;v++;ins(u,v);ins(v,u);
	}
	int cnt=0,s=0;long long ans=0;
	for(int i=1;i<=n;i++)
	{
		if(!t[i])
		{
			Min=1e9;k2=0;
			dfs(i,++s);
			for(int j=1;j<=k2;j++)if(val[c[j]]==Min){ans+=Min;Min=1e9;}else b[++cnt]=val[c[j]];
		}
	}
	if(s==1)return 0*puts("0");
	int lst=(s-1)*2-s;
	sort(b+1,b+cnt+1);
	if(lst>cnt)return 0*puts("Impossible");
	for(int i=1;i<=lst;i++)ans+=b[i];
	printf("%lld",ans);return 0;
 } 