Problem: We have a connected undirected graph with N vertices and M edges. Edge i in this graph (1 \leq i \leq M) connects Vertex U_i and Vertex V_i bidirectionally. We are additionally given N integers D_1, D_2, ..., D_N.

Determine whether the conditions below can be satisfied by assigning a color - white or black - to each vertex and an integer weight between 1 and 10^9 (inclusive) to each edge in this graph. If the answer is yes, find one such assignment of colors and integers, too.

* There is at least one vertex assigned white and at least one vertex assigned black.
* For each vertex v (1 \leq v \leq N), the following holds.
* The minimum cost to travel from Vertex v to a vertex whose color assigned is different from that of Vertex v by traversing the edges is equal to D_v.



Here, the cost of traversing the edges is the sum of the weights of the edges traversed.

Constraints

* 2 \leq N \leq 100,000
* 1 \leq M \leq 200,000
* 1 \leq D_i \leq 10^9
* 1 \leq U_i, V_i \leq N
* The given graph is connected and has no self-loops or multiple edges.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M
D_1 D_2 ... D_N
U_1 V_1
U_2 V_2
\vdots
U_M V_M


Output

If there is no assignment satisfying the conditions, print a single line containing `-1`.

If such an assignment exists, print one such assignment in the following format:


S
C_1
C_2
\vdots
C_M


Here,

* the first line should contain the string S of length N. Its i-th character (1 \leq i \leq N) should be `W` if Vertex i is assigned white and `B` if it is assigned black.
* The (i + 1)-th line (1 \leq i \leq M) should contain the integer weight C_i assigned to Edge i.

Examples

Input

5 5
3 4 3 5 7
1 2
1 3
3 2
4 2
4 5


Output

BWWBB
4
3
1
5
2


Input

5 7
1 2 3 4 5
1 2
1 3
1 4
2 3
2 5
3 5
4 5


Output

-1


Input

4 6
1 1 1 1
1 2
1 3
1 4
2 3
2 4
3 4


Output

BBBW
1
1
1
2
1
1

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 1e5;
const int M = 2 * (int)1e5;
vector<int> conns[N];
int dist[M];
int dep[N];

void dfs(int i, string& col) {
	for (int t : conns[i]) {
		if (dep[t] >= dep[i] && col[t] == '_') {
			col[t] = (col[i] == 'W' ? 'B' : 'W');
			dfs(t, col);
		}
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	int n, m;
	cin >> n >> m;

	vector<pair<int, int>> ds(n);
	for (int i = 0; i < n; ++i) {
		cin >> dep[i];
		ds[i] = {dep[i], i};
	}
	sort(ds.begin(), ds.end());
	int d0 = ds[0].first;

	for (int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		conns[a].push_back(b);
		conns[b].push_back(a);
		dist[i] = max(dep[a], dep[b]);
	}

	bool fail = false;
	string col(n, '_');
	for (auto pr : ds) {
		int i = pr.second;
		if (col[i] == '_') {
			col[i] = 'W';
			dfs(i, col);
		}
		bool works = false;
		for (int t : conns[i]) works |= (dep[t] <= dep[i]);
		fail |= !works;
	}

	if (fail) {
		cout << -1 << '\n';
	} else {
		cout << col << '\n';
		for (int i = 0; i < m; ++i) cout << dist[i] << ' '; cout << '\n';
	}
}