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>
using namespace std;
int n,m,fa[100005],a[100005],b[100005],x,y;
long long s;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
int find(int r){return fa[r]?fa[r]=find(fa[r]):r;}
int main(){
	n=read(),m=read();
	if(m==n-1) return 0*puts("0");
	if(m<n-m-2) return 0*puts("Impossible");
	for(int i=1;i<=n;++i) a[i]=read(),s+=a[i];
	for(int i=1;i<=m;++i){
		x=read(),y=read();
		x=find(x+1),y=find(y+1);
		s-=max(a[x],a[y]);
		b[i]=max(a[x],a[y]);
		if(a[x]>a[y]) fa[x]=y;else fa[y]=x;
	}
	sort(b+1,b+m+1);
	for(int i=1;i<=n-m-2;++i) s+=b[i];
	printf("%lld",s);
}
//既然输入的都是树，并且要求加的边的点权最小，不妨将每一棵树考虑成一棵根节点值最小的树
//显然最优的情况就是把每一棵树的顶点和剩下最小的n-m-2个点连接（根->根->子节点->根->子节点->根...用n-m-2个点串n-m棵树） 
//用并查集维护根最小，先把所有点加入sum，每次加入边的时候找它们所在树的根，把大的点从sum里删除并并入小的点所在的树 
//这样就可以保证每次处理时sum只剩下当前所有树的根节点之和 
//最后剩下的sum一定是所有树的根的和
//排序一下剩下的点，取最小的n-m-2个点计入sum即为答案 