// Original C++ solution (slower):
#include<cstdio>

#include<cstring>

#include<algorithm>

#include<queue>

#define mp(x,y) make_pair(x,y) 

#define ll long long

#define MN 200005

using namespace std;

inline ll in(){

    ll x=0;bool f=0;char c;

    for (;(c=getchar())<'0'||c>'9';f=c=='-');

    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');

    return f?-x:x;

}

typedef pair<ll,int> pr;

//priority_queue<pr,vector<pr>,greater<pr> >q;

priority_queue<ll,vector<ll>,greater<ll> >q;

struct edge{

	int to,nxt;

}e[MN<<1];

ll a[MN],res;

int d[MN],fa[MN],rk[MN],num[MN],h[MN],ch[2][MN];

int n,m,x,y,cnt,tot,fx,fy,rt,ct;

bool vis[MN];

ll mn[MN];bool tmp[MN];

inline int gf(int x){return fa[x]?fa[x]=gf(fa[x]):x;}

inline void ins(int x,int y){

	e[++cnt].to=y;e[cnt].nxt=h[x];h[x]=cnt;

}

inline void dfs(int u){

	mn[tot]=min(mn[tot],a[u]);

	for (int i=h[u];i;i=e[i].nxt){

		int v=e[i].to;

		if (vis[v]) continue;

		vis[v]=1;num[v]=tot;dfs(v);

	}

}

inline int merge(int x,int y){

    if (!x) return y;if (!y) return x;

    if (a[x]>a[y]||(a[x]==a[y]&&x>y)) swap(x,y);

    ch[1][x]=merge(ch[1][x],y);fa[ch[1][x]]=x;

    if (d[ch[0][x]]<d[ch[1][x]]) swap(ch[0][x],ch[1][x]);

    d[x]=d[ch[1][x]]+1;return x;

}

inline int pop(int x){

    a[x]=-1;fa[x]=-1;fa[ch[0][x]]=fa[ch[1][x]]=0;

    return merge(ch[0][x],ch[1][x]);

}

int main()

{

	n=in();m=in();

	memset(mn,0x3f,sizeof(mn));

	for (int i=1;i<=n;++i) a[i]=in();

	for (int i=1;i<=m;++i){

		x=in()+1;y=in()+1;

		ins(x,y);ins(y,x);

	}

	for (int i=1;i<=n;++i)

	if (!vis[i]){

		vis[i]=1;num[i]=(++tot);

		rk[tot]=i;dfs(i);

	}

	if (tot==1) {printf("0");return 0;}

	for (int i=1;i<=tot;++i) res+=1ll*mn[i];

	for (int i=1;i<=n;++i){

		if (!tmp[num[i]]&&a[i]==mn[num[i]]) {tmp[num[i]]=1;continue;}

		else q.push(a[i]);

	}

	while (!q.empty()){

		if (ct==tot-2) break;

		ll x=q.top();q.pop();++ct;res+=1ll*x;

	} 

	if (ct==tot-2) printf("%lld",res);else printf("Impossible");

	/*

	for (int i=1;i<=n;++i){

		if (rk[num[i]]==i) continue;

		fx=gf(i),fy=gf(rk[num[i]]);merge(fx,fy);

	}

	for (int i=1;i<=tot;++i){

		fx=gf(rk[i]);q.push(mp(a[fx],fx));

	}

	while (!q.empty()){

		pr x=q.top();q.pop();

		if (q.empty()) break;pr y=q.top();q.pop();

		res+=1ll*x.first+y.first;++ct;

		fx=gf(x.second),fy=gf(y.second);printf("%d %d\n",fx,fy);

		fx=pop(fx);fy=pop(fy);rt=merge(fx,fy);

		if (rt) q.push(mp(a[rt],rt));

	}*/

//	if (ct+m!=n-1) printf("Impossible");

//	else 

//	printf("%lld",res);

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<algorithm>

using namespace std;

typedef long long ll;

int p[100100];

ll ans;

int n, m;

pair<ll, int> a[100100];

bool group_used[100100];

bool used[100100];

int root(int x) { return x == p[x] ? x : p[x] = root(p[x]); }

int main()

{

	scanf("%d%d", &n, &m);

	for (int i = 1; i <= n; i++)

	{

		ll cost;

		scanf("%lld", &cost);

		a[i] = make_pair(cost, i);

	}

	for (int i = 1; i <= n; i++) p[i] = i;

	for (int i = 1; i <= m; i++)

	{

		int u, v;

		scanf("%d%d", &u, &v);

		u++; v++;

		p[root(u)] = root(v);

	}

	int g = 0;

	for (int i = 1; i <= n; i++) if (p[i] == i) g++;

	// need to group g supernodes into a tree. need to have:

	//   sum of degree = 2(g-1) <= number of nodes = n

	if (2 * (g - 1) > n)

	{

		printf("Impossible\n");

		return 0;

	}

	// select smallest cost from each group first

	// then select smallest from rest

	sort(&a[1], &a[n + 1]);

	int dsum = 0;

	for (int i = 1; i <= n && dsum < 2 * (g - 1); i++)

	{

		int x = a[i].second;

		int r = root(x);

		if (group_used[r] || used[x]) continue;

		group_used[r] = used[x] = true;

		ans += a[i].first;

		dsum++;

	}

	for (int i = 1; i <= n && dsum < 2 * (g - 1); i++)

	{

		int x = a[i].second;

		if (used[x]) continue;

		ans += a[i].first;

		dsum++;

	}

	printf("%lld\n", ans);

	return 0;

}


// Fastest known solution for this problem: