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:
def main():
    import sys
    from collections import deque
    input = sys.stdin.readline

    class UnionFind():
        def __init__(self, n):
            self.n = n
            self.root = [-1] * (n + 1)
            self.rnk = [0] * (n + 1)

        def find_root(self, x):
            if self.root[x] < 0:
                return x
            else:
                self.root[x] = self.find_root(self.root[x])
                return self.root[x]

        def unite(self, x, y):
            x = self.find_root(x)
            y = self.find_root(y)
            if x == y:
                return
            elif self.rnk[x] > self.rnk[y]:
                self.root[x] += self.root[y]
                self.root[y] = x
            else:
                self.root[y] += self.root[x]
                self.root[x] = y
                if self.rnk[x] == self.rnk[y]:
                    self.rnk[y] += 1

        def isSameGroup(self, x, y):
            return self.find_root(x) == self.find_root(y)

        def size(self, x):
            return -self.root[self.find_root(x)]

    N, M = map(int, input().split())
    A = list(map(int, input().split()))
    edge = []
    for _ in range(M):
        u, v = map(int, input().split())
        edge.append((u, v))

    e2i = {}
    for i, e in enumerate(edge):
        u, v = e
        e2i[u*(N+1)+v] = i
    adj = [[] for _ in range(N+1)]
    ok = [0] * (N+1)
    ans = [0] * M
    edge.sort(key=lambda e: max(A[e[0]-1], A[e[1]-1]))
    UF = UnionFind(N+1)
    adj = [[] for _ in range(N+1)]
    for u, v in edge:
        if UF.isSameGroup(u, v):
            continue
        UF.unite(u, v)
        adj[u].append(v)
        adj[v].append(u)
        if A[u-1] == A[v-1]:
            ok[u] = 1
            ok[v] = 1
            ans[e2i[u*(N+1)+v]] = A[u-1]
        elif A[u-1] < A[v-1]:
            ok[v] = 1
            ans[e2i[u*(N+1)+v]] = A[v-1]
        else:
            ok[u] = 1
            ans[e2i[u*(N+1)+v]] = A[u-1]
    if sum(ok) != N:
        print(-1)
        exit()
    que = deque()
    que.append(1)
    seen = [0] * (N+1)
    seen[1] = 1
    while que:
        v = que.popleft()
        for u in adj[v]:
            if seen[u] == 0:
                seen[u] = seen[v] * (-1)
                que.append(u)
    bw = ['B' if seen[i] == 1 else 'W' for i in range(1, N+1)]
    print(''.join(bw))
    for a in ans:
        if a == 0:
            print(10**9)
        else:
            print(a)


if __name__ == '__main__':
    main()