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 = 1e6 + 5;

vector<int> g[N];

int d[N];
int vis[N];
bool ok[N];
int a[N], b[N];

int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; ++i)
    scanf("%d", d+i);
  for (int i = 0; i < m; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u, --v;
    a[i] = u;
    b[i] = v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  vector<int> nodes(n);
  for (int i = 0; i < n; ++i)
    nodes[i] = i;
  sort(nodes.begin(), nodes.end(), [](int l, int r) {
    return d[l] < d[r];
  });
  for (int v : nodes) {
    if (vis[v]) continue;
    queue<int> que;
    vis[v] = 1;
    que.push(v);
    while (!que.empty()) {
      int u = que.front();
      que.pop();
      for (int w : g[u]) {
        if (vis[w] || d[w] < d[u])
          continue;
        vis[w] = -vis[u];
        que.push(w);
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    int u = a[i], v = b[i];
    if (vis[u] != vis[v]) {
      if (d[u] == max(d[u], d[v]))
        ok[u] = 1;
      if (d[v] == max(d[u], d[v]))
        ok[v] = 1;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (!ok[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%c", vis[i] > 0 ? 'B' : 'W');
  }
  printf("\n");
  for (int i = 0; i < m; ++i)
    printf("%d\n", max(d[a[i]], d[b[i]]));
  return 0;
}