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;

#define FOE(i, s, t) for (int i = s; i <= t; i++)
#define FOR(i, s, t) for (int i = s; i < t; i++)
#define FOD(i, s, t) for (int i = s; i >= t; i--)

#define LL long long
#define mp make_pair
#define pb push_back

#define K 600001

int n, m;
int a[K];
vector<int> nxt[K];

LL ret = 0;
int k = 0;

vector<int> sub;
vector<int> poll;

int dfs[K];

void du(int u) {
	dfs[u] = 1;
	sub.pb(a[u]);
	
	for (auto v : nxt[u]) {
		if (!dfs[v]) {
			du(v);
		}	
	}
}

void dd(int u) {
	sub.clear();
	k++;
	
	du(u);	
	
	sort(sub.begin(), sub.end());

	ret += sub[0];
	FOR(i, 1, sub.size()) poll.pb(sub[i]);
}

int main() {
	scanf("%d%d", &n, &m);
	FOE(i, 1, n) scanf("%d", &a[i]);
	FOE(i, 1, m) {
		int u, v; scanf("%d%d", &u, &v);
		u++; v++;
		nxt[u].pb(v);
		nxt[v].pb(u);	
	}
	
	FOE(i, 1, n) if (!dfs[i]) dd(i);
	
	if (k == 1) {
		puts("0");
		return 0;	
	}
	
	int needed = k - 2;
	
	if ((int)poll.size() < needed) {
		puts("Impossible");
		return 0;	
	}
	
	sort(poll.begin(), poll.end());
	
	FOR(i, 0, k - 2) ret += poll[i];
	
	printf("%lld\n", ret); 
}