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;

const int maxn = 1e5, maxm = 4e5, inf = 1e9;
int n, m, d[maxn + 3], tot, ter[maxm + 3], nxt[maxm + 3], wei[maxm + 3], lnk[maxn + 3], to[maxn + 3], id[maxn + 3];
vector<int> S[maxn + 3];
bool vis[maxn + 3];
char s[maxn + 3];

void add(int u, int v) {
	ter[++tot] = v, wei[tot] = inf, nxt[tot] = lnk[u], lnk[u] = tot;
}

void dfs(int u) {
	vis[u] = true;
	for (int i = 0, e, v; i < S[u].size(); i++) {
		e = S[u][i], v = ter[e];
		if (!vis[v]) {
			wei[e] = wei[e & 1 ? e + 1 : e - 1] = d[v];
			s[v] = s[u] ^ 'W' ^ 'B', dfs(v);
		}
	}
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", &d[i]);
	for (int i = 1, u, v; i <= m; i++) {
		scanf("%d %d", &u, &v), add(u, v), add(v, u);
	}
	bool flag = true;
	for (int i = 1; i <= n; i++) {
		pair<int, int> p = make_pair(inf, inf);
		int e = 0;
		for (int j = lnk[i], v; j; j = nxt[j]) {
			v = ter[j];
			if (make_pair(d[v], v) < p) {
				p = make_pair(d[v], v), e = j;
			}
		}
		if (p.first > d[i]) { flag = false; break; }
		to[i] = p.second, id[i] = e, S[to[i]].push_back(e & 1 ? e + 1 : e - 1);
	}
	if (!flag) puts("-1"), exit(0);
	for (int i = 1; i <= n; i++) if (!vis[i] && to[to[i]] == i) s[i] = 'W', dfs(i);
	puts(s + 1);
	for (int i = 1; i <= m; i++) printf("%d\n", wei[i * 2]);
	return 0;
}