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>
#include<iostream>
#include<cstdio>
#define ll long long
#define N 100010
using namespace std;

ll n,m,tot,ans,a[N],father[N],cnt[N];
vector<ll> t,q[N];

inline ll read(){
	ll x=0,tmp=1;
	char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-') tmp=-1;
		ch=getchar();
	}
	while(isdigit(ch)){
		x=(x<<3)+(x<<1)+(ch^48);
		ch=getchar();
	}
	return tmp*x;
}

ll searchfather(ll v){
	if(father[v]==v) return father[v];
	return father[v]=searchfather(father[v]);
}

int main(){
	n=read(); m=read();
	for(ll i=1; i<=n; i++){
		a[i]=read();
		father[i]=i;
	}
	while(m--){
		ll x=read()+1,y=read()+1;
		ll f1=searchfather(x),f2=searchfather(y);
		if(father[f1]!=f2) father[f1]=f2;
	}
	for(ll i=1; i<=n; i++){
		father[i]=searchfather(i);
		if(!cnt[father[i]]) cnt[father[i]]=++tot;
		q[cnt[father[i]]].push_back(a[i]);
	}
	if(tot==1) return printf("0\n")&0;
	for(ll i=1; i<=tot; i++) sort(q[i].begin(),q[i].end());
	for(ll i=1; i<=tot; i++) ans+=q[i][0];
	for(ll i=1; i<=tot; i++){
		for(ll j=1; j<q[i].size(); j++) t.push_back(q[i][j]);
	}
	sort(t.begin(),t.end());
	if(t.size()<tot-2) return printf("Impossible\n")&0;
	for(ll i=0; i<tot-2; i++) ans+=t[i];
	cout<<ans<<endl;
	return 0;
}