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;

vector<int> d, ans, c, par;
vector<pair<int, int>> edge;
vector<vector<int>> g;

void dfs(int x, int last) {
  if (~c[x]) return ;
  int v = edge[par[x]].second;
  if (v == last) {
    c[x] = 0;
    return ;
  }
 
  dfs(v, x);
  ans[par[x] - (par[x]&1)] = d[x];
  c[x] = c[v]^1;
}

int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);

  int n, m;
  cin >> n >> m;
  
  g.resize(n);
  d.resize(n);
  c.resize(n, -1);
  par.resize(n, -1);
  ans.resize(2 * m, 1e9);

  for (int i = 0; i < n; ++i) cin >> d[i];
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    --u; --v;
  
    g[u].push_back(edge.size());
    g[v].push_back(edge.size()+1);
    edge.emplace_back(u, v);
    edge.emplace_back(v, u);
  }

  for (int i = 0; i < n; ++i) {
    int p = -1, val = 1e9 + 5;
    for (auto e : g[i]) {
      int v = edge[e].second;
      if (d[v] < val) {
        val = d[v];
        p = e;
      } else if (d[v] == val) p = min(p, e);
    }
    if (val > d[i]) {
      cout << -1 << '\n';
      return 0;
    }
    par[i] = p;
  }

  for (int i = 0; i < n; ++i) {
    dfs(i, i);
  }
  
  for (int i = 0; i < n; ++i) cout << (c[i] ? 'B' : 'W');
  cout << '\n';
  for (int i = 0; i < m; ++i) cout << ans[2 * i] << '\n';
}