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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct bian
{
	int nxt,to;
}bi[400040];
int n,m,d[100010],head[100010],num,w[200020],p[100010];
char col[100010];
inline void add(int from,int to)
{
	bi[++num]=bian{head[from],to};
	head[from]=num;
}
int cnt;
bool vis[100010];
void dfs(int x,int c)
{
	++cnt;vis[x]=1;col[x]=c?'W':'B';
	for(int i=head[x],y;i;i=bi[i].nxt)
	{
		y=bi[i].to;
		if(d[y]==d[x]&&!vis[y])w[(i+1)/2]=d[x],dfs(y,c^1);
	}
}
void work(int x)
{
	cnt=0;
	dfs(x,0);
	if(cnt==1)col[x]=0;
}
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;++i)scanf("%d",&d[i]),p[i]=i;
	for(int i=1,x,y;i<=m;++i)
	{
		scanf("%d %d",&x,&y);
		add(x,y),add(y,x);
	}
	sort(p+1,p+1+n,[&](int x,int y){return d[x]<d[y];});
	for(int i=1,r;i<=n;i=r+1)
	{
		r=i;
		while(r<n&&d[p[r+1]]==d[p[i]])++r;
		for(int j=i;j<=r;++j)if(!col[p[j]])work(p[j]);
		for(int j=i;j<=r;++j)
			if(!col[p[j]])
			{
				bool flag=0;
				for(int k=head[p[j]];k;k=bi[k].nxt)
					if(d[bi[k].to]<d[p[j]])
					{
						w[(k+1)/2]=d[p[j]]-d[bi[k].to];
						col[p[j]]=col[bi[k].to];
						flag=1;
						break;
					}
				if(!flag)
				{
					printf("-1");
					exit(0);
				}
			}
	}
	for(int i=1;i<=n;++i)putchar(col[i]);putchar('\n');
	for(int i=1;i<=m;++i)
		if(w[i])printf("%d\n",w[i]);
		else printf("%d\n",1000000000);
	return 0;
}