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:
// They can't break me, as long as I know who I am...

#include<bits/stdc++.h>

#define F first
#define S second
#define PB push_back
#define sz(s) int((s).size())
#define bit(n,k) (((n)>>(k))&1)

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

const int maxn = 2e5 + 10, inf = 1e9;

int a[maxn], ans[maxn];
bool col[maxn], mark[maxn];
vector<int> v[maxn], g[maxn], vid[maxn], gid[maxn];
map<int, vector<int> > mp;

void dfs(int u, int w){
    mark[u] = 1;
    for(int i = 0; i < sz(g[u]); i++){
	if(mark[g[u][i]])
	    continue;
	col[ g[u][i] ] = col[u] ^ 1;
	ans[ gid[u][i] ] = w;
	dfs(g[u][i], w);
    }
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie();

    int n ,m;
    cin >> n >> m;
    for(int i = 1; i <= n; i++){
	cin >> a[i];
	mp[a[i]].PB(i);
    }
    for(int i = 0; i < m; i++){
	int x, y;
	cin >> x >> y;
	if(a[x] > a[y])
	    swap(x, y);
	if(a[x] < a[y])
	    v[y].PB(x), vid[y].PB(i);
	else
	    g[x].PB(y), g[y].PB(x), gid[x].PB(i), gid[y].PB(i);
    }
    for(int u = 1; u <= n; u++){
	if(g[u].empty() && v[u].empty())
	    return cout << -1 << endl, 0;
    }

    fill(ans, ans + m, inf);
    
    for(auto it : mp){
	for(int u : it.S){
	    if(mark[u] == 0){
		if(g[u].empty())
		    col[u] = col[ v[u].back() ], ans[ vid[u].back() ] = a[u] - a[ v[u].back() ];
		else
		    dfs(u, it.F);
	    }
	}
    }
    for(int i = 1; i <= n; i++)
	cout << (col[i] ? 'B' : 'W');
    cout<<endl;
    for(int i = 0; i < m; i++)
	cout << ans[i] << "\n";
    return 0;
}
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")