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<vector>
#include<algorithm>
#define INF 2100000000
#define MAXN 100006
using namespace std;
vector<int> road[MAXN];
int v[MAXN],k,minline[MAXN],cnt,n,m,block[MAXN],minone[MAXN],a,b;
long long ans;
bool vis[MAXN];
void DFS(int x,int co)
{
	block[x]=co;
	if(v[minone[co]]>v[x]) minone[co]=x;
	for(int i=0;i<(int)road[x].size();i++)
	{
		int nex=road[x][i];
		if(block[nex]) continue;
		DFS(nex,co);
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d",&v[i]);
	v[0]=INF;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&a,&b);
		a++;
		b++;
		road[a].push_back(b);
		road[b].push_back(a);
	}
	for(int i=1;i<=n;i++)
		if(!block[i]) DFS(i,++k);
	if(k==1)
	{
		printf("0");
		return 0;
	}
	if(k*2-2>n)
	{
		printf("Impossible");
		return 0;
	}
	for(int i=1;i<=k;i++)
	{
		ans+=1ll*v[minone[i]];
		vis[minone[i]]=1;
	}
	for(int i=1;i<=n;i++)
		if(!vis[i]) minline[++cnt]=v[i];
	sort(minline+1,minline+1+cnt);
	for(int i=1;i<=k-2;i++)
		ans+=1ll*minline[i];
	printf("%lld",ans);
}