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;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}

int nextInt() { int x; scanf("%d", &x); return x;}
ll nextLong() { ll x; scanf("%lld", &x); return x;}

const int MAX_N = 112345;
const int MAX_M = 212345;
int D[MAX_N];
int W[MAX_M];
int C[MAX_N];

struct E {
  int id;
  int from, to;
};

vector<E> g[MAX_N];


void dfs(int cur, int next_color, int prev) {
  if (C[cur] != 0) return;
  C[cur] = next_color;
  for (E e : g[cur]) if (e.to != prev && W[e.id] >= 0) {
    dfs(e.to, -next_color, cur);
  }
}

int main2() {
  REP(i, MAX_N) g[i].clear();

  int N = nextInt();
  int M = nextInt();
  REP(i, N) D[i] = nextInt();
  REP(i, M) {
    int a = nextInt() - 1;
    int b = nextInt() - 1;
    g[a].push_back((E){i, a, b});
    g[b].push_back((E){i, b, a});
  }

  CLR(W, -1);


  bool possible = true;
  REP(i, N) {
    sort(ALL(g[i]), [](const E&a, const E&b){
      if (D[a.to] != D[b.to]) return D[a.to] < D[b.to];
      return a.to < b.to;
    });
    E e = g[i][0];
    W[e.id] = D[i];

    if (D[e.to] > D[i]) possible = false;
  }

  if (!possible) {
    cout << -1 << endl;
    return 0;
  }

  CLR(C, 0);

  REP(i, N) {
    dfs(i, 1, -1);
  }
  REP(i, N) cout << (C[i] == 1 ? "B":"W"); cout << endl;
  REP(i, M) {
    if (W[i] < 0) W[i] = 1000000000;
    cout << W[i] << endl;
  }
  return 0;
}

int main() {

#ifdef LOCAL
  for (;!cin.eof();cin>>ws)
#endif
    main2();
  return 0;
}