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;
typedef long long ll;

const int nax = 1e5;
vector<pair<int,int>> node[nax];
vector<int> child[nax];
int weight[nax*2], par[nax];
int col[nax];

void dfs(int p, int c) {
  if (col[p]) return;
  col[p] = c;
  for (int i : child[p])
    dfs(i, 3-c);
}

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> d(n);
  for (int i = 0; i < n; i++) {
    cin >> d[i];
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    node[a].emplace_back(b, i);
    node[b].emplace_back(a, i);
    weight[i] = 1e9;
  }
  for (int i = 0; i < n; i++) {
    par[i] = -1;
    for (auto pa : node[i]) {
      int j = pa.first, ei = pa.second;
      if (d[j] <= d[i]) {
	weight[ei] = d[i];
	par[i] = j;
	if (d[j] == d[i]) {
	  par[j] = i;
	}
      }
    }
    if (par[i] == -1) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++)
    child[par[i]].push_back(i);
  for (int i = 0; i < n; i++) {
    if (par[par[i]] == i) {
      dfs(i, 1);
    }
  }
  for (int i = 0; i < n; i++)
    cout << (col[i] == 2 ? "B" : "W");
  cout << '\n';
  for (int i = 0; i < m; i++)
    cout << weight[i] << '\n';
}