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 <map>
#include <set>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

#define mp make_pair
#define pb push_back
#define x first
#define y second

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;

const int N=100010;

vi e[N];
pii a[N];
bool ok[N];
int col[N];

void dfs(int k,int c)
{
	col[k]=c;
	for (vi::iterator p=e[k].begin();p!=e[k].end();p++)
		if (!col[*p]) dfs(*p,c);
}

int main()
{
	int n,m;scanf("%d%d",&n,&m);
	if (m==n-1) {puts("0");return 0;}
	if (n-1-m>n/2) {puts("Impossible");return 0;}
	for (int i=0;i<n;i++) {int x;scanf("%d",&x);a[i]=mp(x,i);}
	for (int i=0;i<m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		e[x].pb(y);e[y].pb(x);
	}
	int l=0;
	for (int i=0;i<n;i++)
		if (!col[i]) dfs(i,++l);
	sort(a,a+n);
	ll ans=0,res=(n-1-m)*2-(n-m);
	for (int i=0;i<n;i++) if (!ok[col[a[i].y]]) {ok[col[a[i].y]]=1;ans+=a[i].x;} else if (res) {res--;ans+=a[i].x;}
	printf("%lld\n",ans);
	return 0;
}
