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;
typedef long long ll;
typedef pair<int, int> pii;
const int MX=100010, inf=2e9;

int n, m;
pii A[MX];

int U[MX];
int find(int x){ return x==U[x] ? x : U[x]=find(U[x]); }
void unite(int x, int y){ U[find(y)]=find(x); }

int main(){
    ios::sync_with_stdio(0); cin.tie(0);
    cin>>n>>m;
    iota(U, U+n, 0);
    for(int i=0; i<n; i++) cin>>A[i].first, A[i].second=i;
    for(int i=1; i<=m; i++){
        int u,v; cin>>u>>v;
        unite(u,v);
    }
    set<int> S;
    for(int i=0; i<n; i++) S.insert(find(i));
    int sz=S.size(), tot=2*sz-2, cnt[MX]={};
    ll ans=0;
    sort(A, A+n);

    if(sz==1){ cout<<"0\n"; return 0; }
    
    for(int i=0; i<n; i++){
        int x=find(A[i].second);
        if(cnt[x]==0){ ans+=A[i].first, tot--, cnt[x]++, A[i].second=-1; }
    }

    for(int i=0; i<n; i++){
        if(tot==0) break;
        int val, idx; tie(val, idx)=A[i];
        if(idx<0) continue;
        idx=find(idx);
        if(cnt[idx]>=sz-1) continue;
        ans+=val, cnt[idx]++, tot--;
    }
    if(tot>0) cout<<"Impossible\n";
    else cout<<ans<<'\n';
    return 0;
}
