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 N = 100005, M = 200005, inf = 1000000000;

template <class T>
void read (T &x) {
	int sgn = 1;
	char ch;
	x = 0;
	for (ch = getchar(); (ch < '0' || ch > '9') && ch != '-'; ch = getchar()) ;
	if (ch == '-') ch = getchar(), sgn = -1;
	for (; '0' <= ch && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	x *= sgn;
}

template <class T>
void write (T x) {
	if (x < 0) putchar('-'), write(-x);
	else if (x < 10) putchar(x + '0');
	else write(x / 10), putchar(x % 10 + '0');
}

int n, m, fa[N], val[N], d[N], e[M];
bool vis[N], flag = true;

int findroot (int u) {
	if (fa[u] == u) return u;
	int r = findroot(fa[u]);
	assert(fa[fa[u]] == r);
	val[u] ^= val[fa[u]], fa[u] = r;
	return r;
}
bool merge (int u, int v) {
	int ru = findroot(u), rv = findroot(v);
	if (ru == rv) return false;
	fa[ru] = rv, val[ru] = val[u] ^ val[v] ^ 1;
	return true;
}
int main () {
	read(n), read(m);
	for (int i = 1; i <= n; i++) {
		fa[i] = i, val[i] = 0;
		vis[i] = false, read(d[i]);
	}
	for (int i = 1; i <= m; i++) {
		int u, v;
		read(u), read(v), e[i] = inf;
		bool need_add = false;
		if (d[v] <= d[u] && !vis[u]) need_add = true;
		if (d[u] <= d[v] && !vis[v]) need_add = true;
		if (need_add) {
			merge(u, v), e[i] = max(d[u], d[v]);
			if (d[v] <= d[u]) vis[u] = true;
			if (d[u] <= d[v]) vis[v] = true;
		}
	}

	for (int i = 1; i <= n; i++) flag &= vis[i];
	if (flag) {
		for (int i = 1; i <= n; i++) findroot(i), putchar(val[i] ? 'B' : 'W');
		putchar('\n');
		for (int i = 1; i <= m; i++) write(e[i]), putchar('\n');
	}
	else puts("-1");
	return 0;
}