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>
using namespace std;
//#define cerr if (false) cerr
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
	return os << "(" << x.first << "," << x.second << ")";
}
typedef long long ll;
typedef long double ld;
char col[100005];
int D[100005];
vector<int> F[100005], E[100005];
bool used[100005];
void go(int x, int w = 0) {
	if (col[x]) return;
	col[x] = "BW"[w];
	used[x] = true;
	for (int y : F[x]) {
		go(y, 1 - w);
	}
}
int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; ++i) scanf("%d", &D[i]);
	vector<pair<int, int>> edges;
	for (int i = 0; i < m; ++i) {
		int u, v;
		scanf("%d%d", &u, &v);
		--u;
		--v;
		edges.emplace_back(u, v);
		E[u].push_back(v);
		E[v].push_back(u);
	}
	vector<int> nex(n, -1);
	for (int i = 0; i < n; ++i) {
		for (int j : E[i]) {
			if (D[j] <= D[i]) {
				nex[i] = j;
				F[j].push_back(i);
				break;
			}
		}
		if (nex[i] == -1) {
			printf("-1\n");
			return 0;
		}
	}
	for (int i = 0; i < n; ++i) {
		if (used[i]) continue;
		int x = i;
		while (!used[x]) {
			used[x] = true;
			x = nex[x];
		}
		go(x);
	}
	col[n] = '\0';
	printf("%s\n", col);
	for (auto& e : edges) {
		int u, v;
		tie(u, v) = e;
		printf("%d\n", max(D[u], D[v]));
	}
}