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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
#define ll long long
#define ld long double
using namespace std;

int D[100010], ans_e[200010], ans_v[100010], mx=1e9;
bool sast[100010];
pair<int, int> dd[100010];

void dfs(int key, vector<vector<int>> &g){
	for(int i=0; i<g[key].size(); ++i){
		if(ans_v[g[key][i]]==0){
			ans_v[g[key][i]]=3-ans_v[key];
			dfs(g[key][i], g);
		}
	}
}

int main() {
	int N, M;
	cin >> N >> M;
	for(int i=1; i<=N; ++i){
		cin >> D[i];
		dd[i-1]=make_pair(D[i], i);
	}
	sort(dd, dd+N);
	vector<vector<pair<int, int>>> G(N+1);
	for(int i=0; i<M; ++i){
		int u, v;
		cin >> u >> v;
		G[u].emplace_back(make_pair(v, i));
		G[v].emplace_back(make_pair(u, i));
	}
	vector<vector<int>> h(N+1);
	for(int i=0; i<N; ++i){
		int s=dd[i].second;
		if(!sast[s]){
			bool flag=true;
			for(int t=0; t<G[s].size(); ++t){
				if(D[G[s][t].first]<=D[s]){
					ans_e[G[s][t].second]=D[s];
					h[s].emplace_back(G[s][t].first);
					h[G[s][t].first].emplace_back(s);
					sast[s]=sast[G[s][t].first]=1;
					flag=false;
					break;
				}
			}
			if(flag){
				cout << -1 << endl;
				return 0;
			}
		}
	}
	for(int i=1; i<=N; ++i){
		if(ans_v[i]==0){
			ans_v[i]=1;
			dfs(i, h);
		}
	}
	for(int i=1; i<=N; ++i) cout << (ans_v[i]==1 ? 'B' : 'W');
	cout << endl;
	for(int i=0; i<M; ++i){
		if(ans_e[i]==0) cout << mx << endl;
		else cout << ans_e[i] << endl;
	}
	return 0;
}