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 ng() {
    cout << -1 << "\n";
    return 0;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n, m; cin >> n >> m;
    vector<int> d(n);
    vector<pair<int, int>> ds;
    for (int i = 0; i < n; i++) { 
        cin >> d[i];
        ds.emplace_back(d[i], i);
    }
    vector<pair<int, int>> e[n], uv;
    vector<int> c(m, 0), col(n, 0);
    for (int i = 0; i < m; i++) {
        int u, v; cin >> u >> v;
        u--; v--;
        uv.emplace_back(u, v);
        e[u].emplace_back(v, i);
        e[v].emplace_back(u, i);
    }
    sort(ds.begin(), ds.end());
    for (int i = 0; i < n; i++) {
        auto &x = ds[i].second;
        auto &p = ds[i].first;
        bool ok = false;
        for (auto &ed: e[x]) {
            if (d[ed.first] <= p) {
                c[ed.second] = p;
                ok = true;
                break;
            }
        }
        if (!ok) return ng();
    }
    vector<int> f[n], visited(n, 0);
    for (int i = 0; i < m; i++) {
        if (c[i]) {
            auto &u = uv[i].first;
            auto &v = uv[i].second;
            f[u].emplace_back(v);
            f[v].emplace_back(u);
        } else c[i] = 1e9;
    }

    function<void(int, int)> dfs = [&](int curr, int prev) {
        visited[curr] = 1;
        for (auto &to: f[curr]) {
            if (to == prev) continue;
            col[to] = !col[curr];
            dfs(to, curr);
        }
    };

    for (int i = 0; i < n; i++) {
        if (!visited[i]) dfs(i, -1);
    }

    for (int i = 0; i < n; i++) {
        cout << (col[i] ? 'B' : 'W');
    }
    cout << "\n";
    for (int i = 0; i < m; i++) {
        cout << c[i] << "\n";
    }
    return 0;
}