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;
typedef pair <int, int> pii;
typedef long long int LL;
const int MAXN = 100005;

int n, m, val[MAXN], f[MAXN], tmp[MAXN], tot, cnt;
pii minv[MAXN]; bool removed[MAXN];

int find(int x){return f[x] ? f[x] = find(f[x]) : x;}
void merge(int u, int v){
	u = find(u), v = find(v);
	f[v] = u, minv[u] = min(minv[v], minv[u]);
}
int main(){
	int i, u, v; LL sum = 0LL;
	scanf("%d%d", &n, &m);
	for(i = 1; i <= n; ++ i){
		scanf("%d", &val[i]);
		minv[i] = pii(val[i], i);
	}
	for(i = 1; i <= m; ++ i){
		scanf("%d%d", &u, &v);
		merge(u + 1, v + 1);
	}
	tot = (n - m - 1) * 2;
	if(n < tot){
		printf("Impossible\n");
		return 0;
	}
	if(tot == 0){
		printf("0\n");
		return 0;
	}
	for(i = 1; i <= n; ++ i)
		if(find(i) == i){
			sum += minv[i].first;
			removed[minv[i].second] = true;
			-- tot;
		}
	for(i = 1; i <= n; ++ i)
		if(!removed[i])
			tmp[++ cnt] = val[i];
	sort(tmp + 1, tmp + 1 + cnt);
	for(i = 1; i <= tot; ++ i)
		sum += tmp[i];
	cout << sum << endl;
	return 0;
}