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 <algorithm>
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
typedef pair <int, int> pii;
pii a[100010];
vector <int> g[100010];
int k,col[100010];
void dfs(int x)
{
    col[x]=k;
    for (int i=0;i<g[x].size();i++)
        if (!col[g[x][i]]) dfs(g[x][i]);
}
int b[100010];
int main()
{
    int n,m,i;
    scanf("%d%d",&n,&m);
    for (i=0;i<n;i++)
    {
        scanf("%d",&a[i].first);
        a[i].second=i;
    }
    for (i=0;i<m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    for (i=0;i<n;i++)
        if (!col[i])
        {
            k++;
            dfs(i);
        }
    sort(a,a+n);
    int c=k;
    if (2*(k-1)>n)
    {
        printf("Impossible\n");
        return 0;
    }
    long long ans=0;
    int l=1;
    for (i=0;l<=2*k-2;i++)
    {
        int x=col[a[i].second];
        if (b[x]==0)
            c--;
        else
            if (2*k-1-l==c) continue;
        b[x]++;
        ans+=a[i].first;
        l++;
    }
    printf("%lld\n",ans);
    return 0;
}
