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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#define MN 200000
#define MM 200000
#define INF 0x7fffffff
int n,m,a[MN+5];
int hd[MN+5],nxt[MM+5],to[MM+5],rn=0;
int mi1,mi2,lc=0,l[MN+5],lft=2,cnt=0;
long long ans=0;
bool usd[MN+5],usd2[MN+5];
inline void add(int u,int v){
	++rn;
	nxt[rn]=hd[u],hd[u]=rn;
	to[rn]=v;
}
void dfs(int u){
	usd[u]=true;
	if(a[u]<mi1){
		mi1=a[u];
		mi2=u;
	}
	for(int i=hd[u];i!=-1;i=nxt[i])
		if(!usd[to[i]]) dfs(to[i]);
}
int main(){
	memset(hd,0xff,sizeof(hd));
	memset(usd,0,sizeof(usd));
	memset(usd2,0,sizeof(usd2));
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=m;i++){
		int u,v; scanf("%d%d",&u,&v);
		u++,v++; add(u,v),add(v,u);
	}
	for(int i=1;i<=n;i++){
		mi1=INF;
		if(!usd[i]) dfs(i);
		else continue;
		cnt++;
		ans+=mi1; usd2[mi2]=true;
	}
	if(cnt==1) puts("0"),exit(0);
	for(int i=1;i<=n;i++)
		if(!usd2[i]) l[++lc]=a[i];
	if(lc<(n-1-m)*2-cnt) puts("Impossible"),exit(0);
	std::sort(l+1,l+1+lc);
	for(int i=1;i<=(n-1-m)*2-cnt;i++)
		ans+=l[i];
	printf("%lld",ans);
}