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;

int n, m;
int D[100005];
int U[200005];
int V[200005];
vector<int> edge[100005];
int col[100005];

int dfs(int x) {
    int ret = 1;
    for (int i : edge[x]) {
        if (col[i]) continue;
        if (D[i] != D[x]) continue;
        col[i] = 3 - col[x];
        ret += dfs(i);
    }
    return ret;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> D[i];
    }
    for (int i = 1; i <= m; i++) {
        cin >> U[i] >> V[i];
        edge[U[i]].push_back(V[i]);
        edge[V[i]].push_back(U[i]);
    }
    int prv = 2;
    for (int x = 1; x <= n; x++) {
        if (col[x]) continue;
        bool en = false;
        for (int i : edge[x]) {
            if (D[i] < D[x]) {
                en = true;
                break;
            }
        }
        prv = 3 - prv;
        col[x] = prv;
        if (dfs(x) == 1 && !en) {
            cout << -1;
            return 0;
        }
    }
    for (int i = 1; i <= n; i++) {
        cout << "BW"[col[i] - 1];
    }
    cout << '\n';
    for (int i = 1; i <= m; i++) {
        int c1 = col[U[i]], c2 = col[V[i]];
        if (c1 == c2) {
            cout << max(1, abs(D[U[i]] - D[V[i]])) << '\n';
        }
        else {
            cout << max(D[U[i]], D[V[i]]) << '\n';
        }
    }
}