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 <queue>
#include <functional>
#include <vector>
using namespace std;
int main() {
    int n, m; scanf(" %d %d", &n, &m);
    int d[n]; for (int i = 0; i < n; i++) scanf(" %d", &d[i]);
    vector<vector<pair<int, int> > > adj(n);
    for (int i = 0; i < m; i++) {
        int u, v; scanf(" %d %d", &u, &v);
        u--; v--;
        adj[u].push_back({v, i});
        adj[v].push_back({u, i});
    }
    const int WMAX = 1e9;
    int w[m], col[n];
    fill(col, col+n, -1);
    fill(w, w+m, WMAX);

    vector<bool> root(n, false);
    for (int i = 0; i < n; i++) {
        bool eq = false, down = false;
        for (auto e : adj[i]) {
            const int j = e.first;
            if (d[j] < d[i]) { down = true; break; }
            else if (d[j] == d[i]) eq = true;
        }
        if (!down) {
            if (!eq) { puts("-1"); return 0; }
            else root[i] = true;
        }
    }
    function <void(int, int)> process_roots = [&](int i, int c) {
        col[i] = c;
        root[i] = true;
        for (auto e : adj[i]) {
            int j = e.first, k = e.second;
            if (d[j] == d[i] && col[j] == -1) {
                w[k] = d[i];
                process_roots(j, 1^c);
            }
        }
    };
    for (int i = 0; i < n; i++) if (root[i] && col[i] == -1) process_roots(i, 0);

    function <void(int, int)> process_tree = [&](int i, int c) {
        for (auto e : adj[i]) {
            const int j = e.first, k = e.second;
            if (d[j] > d[i] && col[j] == -1) {
                w[k] = col[i] == c ? d[j] - d[i] : d[j];
                col[j] = c;
                process_tree(j, c);
            }
        }
    };
    for (int i = 0; i < n; i++) if (root[i]) process_tree(i, 1^col[i]);

    for (int i = 0; i < n; i++) putchar(col[i] == 0 ? 'B' : 'W');
    putchar('\n');
    for (int i = 0; i < m; i++) printf("%d\n", w[i]);
}