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<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
typedef pair<int,int> pii;
int fa[N],vis[N],mi[N],V[N];
vector<pii >Q(N);
int find(int x)
{
    return x==fa[x]?x:(fa[x]=find(fa[x]));
}
int main()
{
    ios::sync_with_stdio(false);
    int n,m,tn=0;
    cin>>n>>m;
    for(int i=0; i<n; i++)
        cin>>Q[i].first,fa[i]=i,mi[i]=INF;
    for(int i=0,u,v; i<m; i++)
    {
        cin>>u>>v,u=find(u),v=find(v);
        if(u!=v)fa[u]=v;
    }
    for(int i=0; i<n; i++)
    {
        find(i);
        if(!vis[fa[i]])V[tn++]=fa[i];
        vis[fa[i]]=1,Q[i].second=fa[i],mi[fa[i]]=min(Q[i].first,mi[fa[i]]);
    }
    if(tn==1)cout<<0;
    else if((tn-1)*2>n)cout<<"Impossible";
    else
    {
        long long s=0;
        for(int i=0; i<tn; i++)s+=mi[V[i]];
        if(tn!=2)sort(Q.begin(),Q.begin()+n);
        for(int i=0; i<n&&tn!=2; i++)
        {
            if(vis[Q[i].second])
            {
                vis[Q[i].second]=0;
                continue;
            }
            s+=Q[i].first,tn--;
        }
        cout<<s;
    }
    return 0;
}
