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;

#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef complex<double> point;
typedef double DOUBLE;
#define xx real()
#define yy imag()

#define REP(i, a, b) for(int i = (a); i < (int)(b); i++)
#define REPN(i, a, b) for(int i = (a); i <= (int)(b); i++)
#define FA(it, x) for(__typeof((x).begin()) it = (x).begin(); it != (x).end(); it++)
#define SZ(x) (int)(x).size()
#define BE(x) (x).begin(), (x).end()
#define SORT(x) sort(BE(x))
#define _1 first
#define _2 second

#define x1 gray_cat_x1
#define y1 gray_cat_y1

template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

#define file "I1"

const double EPS = 1e-9;
const double PI = acos(-1.);
const ll LL_INF = 1e17 + 10;
const int INF = 1e9 + 10;
const ll MOD = 1e9 + 7;

const int MAXN = 2e5 + 5;
const int MAXW = 1e9;

pair<pii, pii> ee[MAXN];

int v[MAXN];
int w[MAXN];
char col[MAXN];

void solve(){
	int n, m;
	scanf("%d%d", &n, &m);
	REPN(i, 1, n) {
		scanf("%d", &v[i]);
	}
	REP(i, 0, m) {
		int a, b;
		scanf("%d%d", &a, &b);
		ee[i] = mp(mp(max(v[a], v[b]), i), mp(a, b));
	}
	sort(ee, ee + m);
	REP(i, 0, m) {
		int a = ee[i]._2._1, b = ee[i]._2._2;
		int ind = ee[i]._1._2;
		if (!col[a] && !col[b]) {
			if (v[a] != v[b]) {
				printf("-1\n");
				return;
			}
			col[a] = 'B';
			col[b] = 'W';
			w[ind] = v[a];
		} else if (col[a] && col[b]) {
			w[ind] = MAXW;
		} else {
			if (!col[a]) {
				swap(a, b);
			}
			if (v[a] < v[b]) {
				col[b] = col[a];
				w[ind] = v[b] - v[a];
			} else if (v[a] == v[b]) {
				col[b] = 'W' - col[a] + 'B';
				w[ind] = v[a];
			}
		}
	}
	printf("%s\n", col + 1);
	REP(i, 0, m) {
		printf("%d\n", w[i]);
	}
}

int main(){

    //freopen(file".in", "r", stdin); freopen(file".out", "w", stdout);
    int t = 1;
    //cin >> t;
    while(t--){
        solve();    
    }
}