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>
#define ll long long
using namespace std;
inline int read()
{int x=0;
char c=getchar();
while (c<'0'||c>'9') c=getchar();
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x;
}
int n,m,head[100005],nxt[200005],v[200005],tot=0;
int a[100005],minn=1279753568,mpos=-1;
bool vis[100005];
ll ans=0;
inline void add(int a,int b)
{tot++;nxt[tot]=head[a];head[a]=tot;v[tot]=b;}
void dfs(int pos,int f)
{vis[pos]=1;
if (a[pos]<minn) {minn=a[pos];mpos=pos;}
for (int i=head[pos];i;i=nxt[i])
{if (v[i]==f) continue;
dfs(v[i],pos);
}
}
int main (){
	int i,x,y;
	n=read();m=read();
	if ((n-1-m)*2>n) {puts("Impossible");return 0;}
	int o=(n-1-m)*2;
	for (i=1;i<=n;i++) a[i]=read();
	for (i=1;i<=m;i++)
	{x=read()+1;y=read()+1;
	add(x,y);add(y,x);
	}
	for (i=1;i<=n;i++)
	{if ((!vis[i])&&o)
	{minn=1279753568;mpos=-1;
	dfs(i,0);
	ans+=minn;o--;
	a[mpos]=1279753568;
	}
	}
	sort(a+1,a+n+1);
	for (i=1;i<=o;i++) ans+=a[i];
	printf ("%lld\n",ans);
	return 0;
}
	
	
	
	
