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 f first
#define s second
#define pb push_back
#define mp make_pair
 
using namespace std;
 
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
 
const int N = (int)2e5 + 123, inf = 1e9, mod = 1e9 + 7;
const ll INF = 1e18;
 
vector< int > g[N];
bool ok[N];
int col[N], d[N], ans[N];
int n, m, s[N], t[N];
void f(int v){
	for(auto i : g[v]){
		int to = s[i] ^ t[i] ^ v;
		if(col[to] == -1){
			col[to] = col[v] ^ 1;
			ans[i] = d[v];
			f(to);
		}
	}
}
int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++)
		scanf("%d", &d[i]);
	for(int i = 1; i <= m; i++){
		scanf("%d%d", &s[i], &t[i]);
		if(d[s[i]] == d[t[i]]){
			g[s[i]].pb(i);
			g[t[i]].pb(i);
			ans[i] = d[s[i]];
			ok[s[i]] = ok[t[i]] = 1;
		}else{
			if(d[s[i]] < d[t[i]])	
				ok[t[i]] = 1;
			else
				ok[s[i]] = 1;
		}
	}   	
	for(int i = 1; i <= n; i++)
		if(!ok[i]){
			printf("-1");
			return 0;
		}

	for(int i = 1; i <= n; i++)
		col[i] = -1;

	for(int i = 1; i <= n; i++){
		if(col[i] == -1){
			col[i] = 0;
			f(i);
		}
	}
	for(int i = 1; i <= n; i++)
		if(col[i] == -1)
			col[i] = 0;
	for(int i = 1; i <= m; i++){
		if(d[s[i]] != d[t[i]]){
			if(col[s[i]] == col[t[i]])
				ans[i] = abs(d[s[i]] - d[t[i]]);
			else
				ans[i] = max(d[s[i]], d[t[i]]);
		}
	}
	for(int i = 1; i <= n; i++)
		printf("%c", (col[i] ? 'B' : 'W'));
	printf("\n");
	for(int i = 1; i <= m; i++)
		printf("%d\n", ans[i]);
}