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 <bits/stdc++.h>

#define endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < n; i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb(x) push_back(x)

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int MX = 200005;
int n, m, d[MX], col[MX], w[MX], bs[MX], q[MX];
vii adj[MX];

void no () {
	cout << -1 << endl;
	exit(0);
}

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin >> n >> m;
	
	for (int i = 1; i <= n; i++)
		cin >> d[i];
	
	forn (i, m) {
		int u, v;
		cin >> u >> v;
		adj[u].emplace_back(v, i);
		adj[v].emplace_back(u, i);
	}

	iota(q, q + n, 1);
	sort(q, q + n, [&] (int i, int j) {
		return d[i] < d[j];
	});

	forn (_, n) {
		int u = q[_];
		if (bs[u]) continue;

		int same = -1, prep = -1;
		for (ii &v : adj[u]) {
			if (bs[v.fi]) prep = v.fi;
			if (d[u] == d[v.fi]) same = v.fi;
		}

		if (prep != -1) {
			col[u] = 1 - col[prep];
			
			for (ii &v : adj[u]) {
				if (v.fi == prep) w[v.se] = d[u];
				else if (bs[v.fi]) w[v.se] = 1e9;
			}
		} else if (same != -1) {
			col[u] = 1;
			
			for (ii &v : adj[u]) {
				if (v.fi == same) w[v.se] = d[u];
				else if (bs[v.fi]) w[v.se] = 1e9;
			}

			for (ii &v : adj[same]) {
				if (bs[v.fi]) w[v.se] = 1e9;
			}
			
			bs[same] = 1;
		} else no();

		bs[u] = 1;
	}

	for (int i = 1; i <= n; i++)
		cout << "WB"[col[i]];
	cout << endl;

	forn (i, m) cout << w[i] << endl;

	return 0;
}