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 PYTHON3):

Solution:
from collections import defaultdict
import sys

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def MI1(): return map(int1, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]

def main():
    def impossible():
        for u in range(n):
            du = dd[u]
            for v in to[u]:
                if dd[v] <= du: break
            else:
                return True
        return False

    def dfs(u=0,pu=-1,c=0):
        color[u]="B" if c else "W"
        for v in to2[u]:
            if v==pu:continue
            dfs(v,u,1-c)

    inf = 10 ** 9
    n, m = MI()
    dd = LI()
    to = defaultdict(list)
    uvtoi = {}
    for i in range(m):
        u, v = MI1()
        uvtoi[u, v] = uvtoi[v, u] = i
        to[u].append(v)
        to[v].append(u)
    if impossible():
        print(-1)
        exit()
    cost = [inf] * m
    to2 = defaultdict(set)
    for u in range(n):
        mn = (inf, inf, inf)
        for v in to[u]:
            t = (dd[v], uvtoi[u, v], v)
            if t < mn: mn = t
        v = mn[2]
        cost[mn[1]] = dd[u]
        to2[u].add(v)
        to2[v].add(u)
    color=["N"]*n
    for u in range(n):
        if color[u]!="N":continue
        dfs(u)
    print("".join(color))
    print(*cost,sep="\n")

main()