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 vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pi;
typedef vector<pi> vpi;
typedef vector<vpi> vvpi;

vi D;
vvpi G;
vi vist, C;

void dfs(int x, int val, int col) {
  vist[x] = col;
  for (auto it : G[x]) {
    int y = it.first;
    if (D[y] == val and vist[y] == -1) {
      C[it.second] = val;
      dfs(y, val, col^1);
    }
  }
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n , m;
  cin >> n >> m;
  D = vi(n);
  G = vvpi(n);
  vpi V(n);
  for (int i = 0; i < n; ++i) {
    cin >> D[i];
    V[i] = pi(D[i], i);
  }
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    --x; --y;
    G[x].push_back(pi(y,i));
    G[y].push_back(pi(x,i));
  }
  sort(V.begin(), V.end());
  bool pots = true;
  C = vi(m, -1);
  vist = vi(n, -1);
  for (int i = 0; i < n; ++i) {
    int x = V[i].second;
    if (vist[x] != -1) continue;
    int minval = D[x] + 1;
    int minar = -1;
    int mincol = -1;
    int contt = 0;
    for (auto it : G[x]) {
      int y = it.first;
      if (D[y] < minval) {
        minar = it.second;
        minval = D[y];
        mincol = vist[y];
      }
      if (D[y] == D[x]) ++contt;
    }
    if (minval > D[x]) {
      pots = false;
      break;
    }
    else if (contt) {
      dfs(x, D[x], 0);
    }
    else {
      C[minar] = D[x];
      vist[x] = mincol^1;
    }
  }
  if (!pots) {
    cout << -1 << '\n';
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    if (vist[i]) cout << 'B';
    else cout << 'W';
  }
  cout << '\n';
  for (int i = 0; i < m; ++i) {
    if (C[i] == -1) cout << 1000000000 << '\n';
    else cout << C[i] << '\n';
  }

}