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;
typedef pair <int , int> pii;

const int N = 2e5+10 , mod = 1e9+7;

int32_t main(){
   ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
   int n , m;
   cin >> n >> m;
   vector <pii> vec;
   for(int i = 0 ; i < n ; i++){
      int x;
      cin >> x;
      vec.push_back({x , i});
   }
   sort(vec.begin() , vec.end());
   int mn[n];
   memset(mn , 63 , sizeof(mn));
   vector <pii> g[n];
   for(int i = 0 ; i < m ; i++){
      int u , v;
      cin >> u >> v;
      u--; v--;
      g[u].push_back({v , i});
      g[v].push_back({u , i});
   }
   int ans[m];
   bool mark[n];
   memset(mark , 0 , sizeof(mark));
   int col[n];
   memset(col , 0 , sizeof(col));
   for(pii x : vec){
      int v = x.second;
      for(pii y : g[v]){
         int u = y.first;
         if(mark[u]){
            if(mn[v] < mod){
               if(mn[u] > mod)
                  col[u] = 1 - col[v];
            }
            else{
               col[v] = 1 - col[u];
            }
            mn[u] = min(mn[u] , x.first);
            mn[v] = min(mn[v] , x.first);
            ans[y.second] = x.first;
         }
      }
      mark[x.second] = true;
   }
   for(pii x : vec)
      if(mn[x.second] != x.first)
         return cout << -1 , 0;
   for(int i = 0 ; i < n ; i++)
      cout << ((col[i] == 0) ? "B" : "W");
   cout << "\n";
   for(int i = 0 ; i < m ; i++)
      cout << ans[i] << "\n";
   return 0;
}