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 rep(i,a,b) for (int i=(a); i<=(b); i++)
#define per(i,a,b) for (int i=(a); i>=(b); i--)
using namespace std;

const int maxn = 100005;
int a[maxn], vis[maxn], n, m, x, y, mn, cur;
priority_queue<int, vector<int>, greater<int> > Q;
vector<int> e[maxn];
long long ans;

void dfs(int u) {
	int sz = e[u].size();
	if (a[u] < mn) mn = a[u], cur = u;
	vis[u] = 1;
	if (!sz) return;
	rep (i, 0, sz - 1)
		if (!vis[e[u][i]]) dfs(e[u][i]);
}

int main() {
	scanf("%d%d", &n, &m);
	if (n < 2 * (n - m - 1)) 
		return puts("Impossible"), 0;
	if (n - 1 == m)
		return puts("0"), 0;
	rep (i, 1, n) scanf("%d", &a[i]);
	rep (i, 1, m) {
		scanf("%d%d", &x, &y); x++; y++;
		e[x].push_back(y); e[y].push_back(x);
	}
	rep (i, 1, n)
		if (!vis[i]) {
			mn = 2e9;
			dfs(i);
			ans += mn;
			a[cur] = 0;
		}
	rep (i, 1, n)
		if (a[i]) Q.push(a[i]);
	m = 2 * (n - m - 1) - (n - m);
	while (m-- > 0) {
		ans += Q.top();
		Q.pop();
	}
	cout << ans << endl;
	return 0;
}
