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:
//
// Created by munaiyi on 2020/3/19.
//

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e5 + 5;
int h[maxn], mem;
struct {
    int v, next, id;
} e[maxn << 2];

void add(int u, int v, int id) {
    e[mem].v = v;
    e[mem].id = id;
    e[mem].next = h[u];
    h[u] = mem++;
}

int vis[maxn], d[maxn];
pair<int, int> dd[maxn];
int ans[maxn << 1];

int main() {
    memset(h, -1, sizeof(h));
    memset(vis, -1, sizeof(vis));
    memset(ans, -1, sizeof(ans));
    mem = 0;
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &d[i]);
        dd[i].first = d[i];
        dd[i].second = i;
    }
    sort(dd + 1, dd + 1 + n);
    for (int i = 0; i < m; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        add(u, v, i);
        add(v, u, i);
    }
    for (int i = 1; i <= n; i++) {
        if (vis[dd[i].second] + 1) continue;
        int uv = -1, id = 0;
        for (int j = h[dd[i].second]; j + 1; j = e[j].next) {
            int v = e[j].v;
            if (vis[v] + 1 && d[v] < d[dd[i].second]) {
                uv = v;
                id = e[j].id;
                break;
            } else if (d[v] == d[dd[i].second]) {
                uv = v;
                id = e[j].id;
            }
        }
        if (uv == -1) {
            printf("-1\n");
            return 0;
        }
        if (d[uv] < d[dd[i].second]) {
            vis[dd[i].second] = vis[uv];
            ans[id] = d[dd[i].second] - d[uv];
        } else {
            if (vis[uv] == -1) {
                vis[uv] = 0;
            }
            vis[dd[i].second] = vis[uv] ^ 1;
            ans[id] = d[dd[i].second];
        }
    }
    for (int i = 1; i <= n; i++) {
        putchar(vis[i] ? 'B' : 'W');
    }
    putchar('\n');
    for (int i = 0; i < m; i++) {
        printf("%d\n", ans[i] == -1 ? 1000000000 : ans[i]);
    }
    return 0;
}