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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long LL;
const int N = 100050;

LL ans = 0;
int t[N], n, m, bj[N];
int mn[N], mnw[N], a[N];
struct node{
	int loc, val;
	inline bool operator < (const node &b)const{
		return val < b.val;
	}
}W[N];
int path(int x){
	if(!t[x])return x;
	return t[x] = path(t[x]);
}
int main()
{

	int i, j, k;
	scanf("%d%d", &n, &m);
	if(2 * (n - m - 1) > n){
		puts("Impossible");
		return 0;
	}
	if(n == m + 1){
		puts("0");
		return 0;
	}
	for(i = 1; i <= n; i ++)
		scanf("%d", &a[i]);
	for(i = 1; i <= m; i ++){
		scanf("%d%d", &j, &k);
		t[path(j + 1)] = path(k + 1);
	}
	for(i = 1; i <= n; i ++)
		mn[i] = 2100000000;
	for(i = 1; i <= n; i ++){
		if(mn[path(i)] > a[i]){
			if(mn[path(i)] != 2100000000)
				ans -= mn[path(i)], bj[mnw[path(i)]] = 0;
			ans += a[i], bj[i] = 1, mnw[path(i)] = i, mn[path(i)] = a[i];
		}
		W[i].loc = i, W[i].val = a[i];
	}
	sort(W + 1, W + n + 1);
	for(i = 1, j = 0; i <= n && j < n - m - 2; i ++)
		if(bj[W[i].loc] == 0)
			j ++, ans += W[i].val;
	printf("%lld", ans);

	return 0;
}
