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 fo(i,a,b) for(int i=(a);i<=(b);++i)
#define rv(i,a,b) for(int i=(a);i>=(b);--i)
#define P pair<int,int>
#define pb push_back
#define lb(x) (x&(-x))
#define ALL(x) x.begin(),x.end()
#define SZ(x) ((int)x.size())
using namespace std;
typedef long long LL;
const int M = 220000;
const int inf = 1e9;
int c[M]={0};
P p[M];
vector<P> g[M];
int n,m,k;
int dd[M];
char s[M];
bool ok(int i){
	return s[i]=='W'||s[i]=='B';
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>m;
	int u,v;
	fo(i,1,n){
		cin>>p[i].first;
		dd[i]=p[i].first;
		p[i].second=i;
	}
	sort(p+1,p+1+n);
	fo(i,1,m){
		cin>>u>>v;
		g[u].push_back({v,i});
		g[v].push_back({u,i});
	}
	int flag=1;
	fo(i,1,n){
		int id=p[i].second;
		if(ok(id)) continue;
		int d=p[i].first;
		bool done=0;
		for(P to:g[id]){
			if(ok(to.first)){
				c[to.second]=d;
				s[id]=(s[to.first]=='W'?'B':'W');
				done=1;
				break;
			}
		}
		if(!done){
			for(P to:g[id]){
				if(dd[to.first]==d){
					done=1;
					s[id]='W';
					s[to.first]='B';
					c[to.second]=d;
				}
			}
		}
		if(!done){
			flag=0;
			break;
		}
	}
	if(flag){
		s[n+1]='\0';
		cout<<s+1<<endl;
		fo(i,1,m){
			cout<<(c[i]?c[i]:inf)<<endl;
		}
	}
	else cout<<-1<<endl;
	return 0;
}