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<queue>
typedef long long ll;
using namespace std;
vector<int> g[100000];
priority_queue<ll,vector<ll>,greater<ll> > q;
int n,m,x,y,visit[100000],cnt;
ll a[100000];

ll dfs(int x,int y,ll z){
    if(z>a[x]){
        q.push(z);
        z=a[x];
    }
    else{
        q.push(a[x]);
    }
    visit[x]=1;
    for(int i=0;i<g[x].size();i++){
        if(g[x][i]!=y){
            z=dfs(g[x][i],x,z);
        }
    }
    return z;
}

int main(){
    ll ans=0;
    scanf("%d %d",&n,&m);
    if(n<2*(n-1-m)){
        printf("Impossible\n");
        return 0;
    }
    else if(n-1==m){
        printf("0\n");
        return 0;
    }
    for(int i=0;i<n;i++){
        scanf("%lld",&a[i]);
    }
    for(int i=0;i<m;i++){
        scanf("%d %d",&x,&y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    for(int i=0;i<n;i++){
        if(visit[i]==0){
            ans+=dfs(i,-1,10000000000);
            cnt++;
        }
    }
    while(2*(n-1)>2*m+cnt){
        ans+=q.top();
        q.pop();
        cnt++;
    }
    printf("%lld\n",ans);
}