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;
const int N = 222222;
int a[N], vst[N], ans[N], col[N];
vector<pair<int, int> > e[N];
int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for(int i(1); i <= n; i++) {
		scanf("%d", &a[i]);
	}
	for(int i(1); i <= m; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		e[x].push_back({y, i});
		e[y].push_back({x, i});
	}
	set<pair<int, int> > st;
	memset(col, -1, sizeof(col));
	for(int i(1); i <= n; i++) {
		for(pair<int, int> tmp : e[i]) {
			int y = tmp.first;
			if(a[i] == a[y]) {
				int v = i;
				vst[i] = vst[y] = 1;
				if(col[v] == -1 && col[y] == -1) {
					col[v] = 0; col[y] = 1;
				}else if(col[v] == -1) {
					col[v] = col[y] ^ 1;
				}else if(col[y] == -1) {
					col[y] = col[v] ^ 1;
				}
				ans[tmp.second] = a[i];
				st.insert({a[i], i});
				st.insert({a[y], y});
			}
		}
	}
	for(int i(1); i <= n; i++) {
		if(st.empty()) {
			printf("-1\n");
			exit(0);
		}
		int v = st.begin()->second;
		//printf("v = %d\n", v);
		st.erase(st.begin());
		for(pair<int, int> tmp : e[v]) {
			int y = tmp.first;
			if(!vst[y] && a[y] >= a[v]) {
				st.insert({a[y], y});
				col[y] = col[v] ^ 1;
				vst[y] = 1;
				ans[tmp.second] = a[y];
			}
		}
	}
	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] ? ans[i] : (int)1e9);
}