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 <vector>
#include <set>
using namespace std;

typedef long long int lli;

const lli maxn = lli(1e5)+5, inf = lli(1e9)+5;

lli A[maxn], vis[maxn];
pair<lli, lli> val[maxn];
vector<lli> graph[maxn];

void dfs(lli node, lli comp)
{
	vis[node] = 1;
	val[comp] = min(val[comp], {A[node], node});
	for(auto it: graph[node])
	{
		if(!vis[it]) dfs(it, comp);
	}
}

int main(void)
{
	lli n, m, x, y;
	scanf("%lld%lld", &n, &m);
	for(lli i = 0;i < n;i++) scanf("%lld", &A[i]);
	for(lli i = 0;i <= n;i++) val[i] = {inf, inf};

	while(m--)
	{
		scanf("%lld%lld", &x, &y);
		graph[x].push_back(y), graph[y].push_back(x);
	}

	lli cnt = 0, res = 0;
	for(lli i = 0;i < n;i++)
	{
		if(!vis[i]) dfs(i, cnt++);
	}

	if(cnt == 1)
	{
		printf("0\n");
		return 0;
	}

	set<pair<lli, lli>> S;
	for(lli i = 0;i < n;i++) S.insert({A[i], i});

	for(lli i = 0;i < cnt;i++) res += val[i].first, S.erase(val[i]);

	cnt -= 2;
	while(cnt--)
	{
		if(S.empty())
		{
			printf("Impossible\n");
			exit(0);
		}
		res += S.begin()->first;
		S.erase(S.begin());
	}
	printf("%lld\n", res);
}