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;

pair<int, int> cost[100001];
vector<int> g[100001];
set<int> st;

int qclr;
int clr[100001];
int vis[100001];

void dfs(int me){
    vis[me] = 1;
    for(auto &nxt: g[me]){
        if(vis[nxt]) continue;
        clr[nxt] = clr[me];
        dfs(nxt);
    }
}

int main(){
    int n,m;
    setbuf(stdout, NULL);
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        cost[i].second = i;
        scanf("%d",&(cost[i].first));
        clr[i] = i;
    }
    for(int i=0;i<m;i++){
        int aa, bb;
        scanf("%d%d",&aa,&bb);
        g[aa].push_back(bb);
        g[bb].push_back(aa);
    }
    for(int i=0;i<n;i++){
        if(vis[i]) continue;
        dfs(i);
    }
    sort(cost, cost + n);
    long long ans = 0;
    int lft = (n - m - 2);
    if(lft >= 0){    
        for(int i=0;i<n;i++){
            if(st.find(clr[cost[i].second]) == st.end()){
                ans += cost[i].first;
                st.insert(clr[cost[i].second]);
            }else if(lft > 0){
                lft--;
                ans += cost[i].first;
            }
        }
    }
    if(lft <= 0) printf("%lld\n", ans);
    else printf("Impossible\n");
}
