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<algorithm>
using namespace std;
typedef long long ll;
const int maxn=100000,inf=0x3f3f3f3f;
int n,m,fa[maxn+10],minv[maxn+10],minid[maxn+10],id[maxn+10];
bool ban[maxn+10]; int a[maxn+10]; ll ans;
int getf(int x){return fa[x]==x?x:fa[x]=getf(fa[x]);}
bool cmp(int x,int y){return a[x]<a[y];}
int main(){
	scanf("%d%d",&n,&m); int ndd=(n-1-m)*2;
	if(ndd>n){
		printf("Impossible"); return 0;
	}
	for(int i=1;i<=n;++i){
		scanf("%d",&a[i]); fa[i]=id[i]=i; minv[i]=inf;
	}
	for(int i=1;i<=m;++i){
		int l,r; scanf("%d%d",&l,&r); ++l; ++r;
		fa[getf(l)]=getf(r);
	}
	for(int i=1;i<=n;++i){
		int p=getf(i);
		if(a[i]<minv[p]) minv[p]=a[i],minid[p]=i;
	}
	int cnt=0; 
	for(int i=1;i<=n;++i) if(getf(i)==i){
		++cnt;
		ans+=minv[i]; ban[minid[i]]=1; --ndd; 
	}
	if(cnt==1){
		printf("0"); return 0;
	}
	sort(id+1,id+n+1,cmp);
	for(int i=1,j=1;i<=ndd;++i){
		for(;ban[id[j]];j++); ans+=a[id[j++]];
	}
	printf("%lld",ans);
}