//看看会不会爆int!数组会不会少了一维！

//取物问题一定要小心先手胜利的条件

#include <bits/stdc++.h>

using namespace std;

#pragma comment(linker,"/STACK:102400000,102400000")

#define LL long long

#define ALL(a) a.begin(), a.end()

#define pb push_back

#define mk make_pair

#define fi first

#define se second

#define haha printf("haha\n")

const int maxn = 1e5 + ;

struct TwoSat{

    int n, c, s[maxn * ];

    bool mark[maxn << ];

    vector<int> G[maxn << ];

    void init(int n){

        this->n = n;

        for (int i = ; i <  * n; i++) G[i].clear();

        memset(mark, false, sizeof(mark));

    }

    void add_edge(int x, int xval, int y, int yval){

        x = x *  + xval, y = y *  + yval;

        G[x ^ ].pb(y), G[y ^ ].pb(x);

    }

    bool dfs(int x){

        if (mark[x ^ ]) return false;

        if (mark[x]) return true;

        mark[x] = true;

        s[c++] = x;

        for (int i = ; i < G[x].size(); i++){

            if (!dfs(G[x][i])) return false;

        }

        return true;

    }

    bool solve(){

        for (int i = ; i < n * ; i += ){

            if (!mark[i] && !mark[i + ]){

                c = ;

                if (!dfs(i)){

                    while (c) mark[s[--c]] = false;

                    if (!dfs(i + )) return false;

                }

            }

        }

        return true;

    }

};

TwoSat sat2;

int n, m;

int age[maxn];

int main(){

    while (scanf("%d%d", &n, &m) && (n + m) > ){

        sat2.init(n);

        int tot = ;

        for (int i = ; i < n; i++){

            scanf("%d", age + i);

            tot += age[i];

        }

/*

假设A和B都是1，C是0

if两者属于同一个任务，那么，两者不能分配到同一个任务，

所以两者不能同时为true和同时为false

if两者属于不同的任务，那么两者不能同时为false

*/

        for (int i = ; i < m; i++){

            int u, v; scanf("%d%d", &u, &v);

            u--, v--;

            if (u == v) continue;

            int tyu = age[u] * n < tot, tyv = age[v] * n < tot;

            sat2.add_edge(u, , v, );

            if (tyu == tyv)

                sat2.add_edge(u, , v, );

        }

        if (!sat2.solve()) puts("No solution.");

        else {

            for (int i = ; i < n; i++){

                if (sat2.mark[i * ]) puts("C");

                if (sat2.mark[i *  + ]) {

                    if (age[i] * n < tot) puts("B");

                    else puts("A");

                }

            }

        }

    }

    return ;

}