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>

#define ll long long
#define fornum(A,B,C) for(A=B;A<C;++A)
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std;

/////////////////////////////////////////////////////

ll N,M;
ll D[101010];
vector<pair<ll,pll>> uvv[101010];
ll ans[101010],ans2[202020];
vector<ll> sv[101010];
ll i, j;

//Union木
struct UnionFind{
    vector<int> nxt;
    void init(int x){
        nxt.clear();
        nxt.resize(x);
        for (int i = 0; i < x;i++){
            nxt[i] = i;
        }
    }
    int find(int x){
        if(nxt[x]==x)
            return x;
        return nxt[x] =find(nxt[x]);
    }
    inline void unite(int x,int y){
        nxt[find(y)] = find(x);
    }
};

void no(){
    printf("-1");
    exit(0);
}
void dfs(ll a,ll c){
    ll i = 0;
    ans[a] = c + 1;
    fornum(i,0,sv[a].size()){
        dfs(sv[a][i], c ^ 1);
    }
}

int main(){
    scanf("%lld%lld",&N,&M);
    fornum(i,0,N){
        scanf("%lld", &D[i + 1]);
    }


    fornum(i,0,M){
        ll u, v;
        scanf("%lld%lld", &u, &v);
        uvv[u].push_back({D[v], {v, i}});
        uvv[v].push_back({D[u], {u, i}});
    }

    fornum(i,0,N){
        sort(uvv[i + 1].begin(), uvv[i + 1].end());
        auto uv = uvv[i + 1][0];
        if(D[i+1]<uv.first){
            no();
        }
        if(ans2[uv.second.second]!=0&&ans2[uv.second.second]!=D[i+1]){
            no();
        }
        if(ans2[uv.second.second]==0){
            ans2[uv.second.second] = D[i + 1];
            sv[uv.second.first].push_back(i + 1);
            ans[i + 1] = 1;
        }
    }
    fornum(i,0,N){
        if (ans[i + 1] != 0)
            continue;
        dfs(i + 1, 0);
    }
    fornum(i,0,N){
        if(ans[i+1]==1){
            printf("B");
        }else{
            printf("W");
        }
    }
    fornum(i,0,M){
        printf("\n%lld", ans2[i] == 0 ? (ll)1e9 : ans2[i]);
    }
    

    
    return 0;
}