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>

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define ar array

using namespace std;
typedef long long ll;
typedef pair<int, int> ii;

const int maxn = 2e5 + 5;

vector<int> gr[maxn], tr[maxn];
int N, M, col[maxn];
int U[maxn], V[maxn];
int w[maxn], d[maxn];
bool in[maxn];
int p[maxn];

void dfs(int u)
{
  for(int v : tr[u]) if(col[v] == -1){
    col[v] = col[u] ^ 1;
    dfs(v);
  }
}

signed main(void)
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  #ifdef LOCAL
    freopen("A.INP", "r", stdin);
    freopen("A.OUT", "w", stdout);
  #endif // LOCAL
  cin >> N >> M;
  for(int i = 1; i <= N; ++i)
    cin >> d[i];
  for(int i = 1; i <= M; ++i){
    cin >> U[i] >> V[i];
    gr[U[i]].eb(i); gr[V[i]].eb(i);
  }
  fill(w + 1, w + 1 + M, 1e9);
  for(int i = 1; i <= N; ++i){
    p[i] = -1;
    for(int id : gr[i]){
      int v = U[id] + V[id] - i;
      if(p[i] == -1 || d[p[i]] > d[v]){
        p[i] = v;
      }
    }
    if(d[i] < d[p[i]]){
      cout << -1;
      return 0;
    }
    tr[i].eb(p[i]);
    tr[p[i]].eb(i);
    for(int id : gr[i]){
      int v = U[id] + V[id] - i;
      if(v == p[i]){
        w[id] = d[i];
        break;
      }
    }
  }
  fill(col + 1, col + 1 + N, -1);
  for(int i = 1; i <= N; ++i){
    if(col[i] == -1){
      col[i] = 0;
      dfs(i);
    }
  }
  for(int i = 1; i <= N; ++i){
    cout << (col[i] ? 'B' : 'W');
  }
  cout << '\n';
  for(int i = 1; i <= M; ++i)
    cout << w[i] << '\n';
}