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;
int ufd[101010];
int Find(int a)
{
    if(a==ufd[a]) return a;
    return ufd[a] = Find(ufd[a]);
}
void Union(int a, int b)
{
    ufd[Find(a)] = Find(b);
}
int N, M;
int a[101010];
vector<pair<int, int> > V;
priority_queue<int> Q[101010];
int main()
{
    scanf("%d%d", &N, &M);
    long long ans = 0;
    for(int i=0; i<N; ++i)
    {
        scanf("%d", a+i);
        V.emplace_back(a[i], i);
        ans += a[i];
        ufd[i] = i;
    }
    sort(V.rbegin(), V.rend());
    for(int i=0; i<M; ++i)
    {
        int x, y; scanf("%d%d", &x, &y);
        Union(x, y);
    }
    if(N==M+1){ puts("0"); return 0;}
    if(N>2*M+2){puts("Impossible"); return 0;}
    int torm = 2*M+2-N;
    for(int i=0; i<N; ++i)
        Q[Find(i)].push(a[i]);
    for(auto x: V)
    {
        if(torm == 0) break;
        if(Q[Find(x.second)].size() > 1)
        {
            //printf("%d %d\n", Q[Find(x.second)].top(), Find(x.second));
            ans -= Q[Find(x.second)].top(); Q[Find(x.second)].pop();
            --torm;
        }
    }
    printf("%lld\n", ans);
}



























