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// {{{
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define FOR(i, b, e) for (int i = (b); i <= (e); i++)
#define FORD(i, b, e) for (int i = (e); i >= (b); i--)
#define MP make_pair
#define FS first
#define ND second
#define PB push_back
#define ALL(x) x.begin(), x.end()

using namespace std;

using LL = long long;
using LD = long double;
using PII = pair<int,int>;
using PLL = pair<LL,LL>;
using PLD = pair<LD,LD>;
using VI = vector<int>;
using VLL = vector<LL>;
using VLD = vector<LD>;
using VB = vector<bool>;
using VS = vector<string>;
template<class T>
using V = vector<T>;
template<class T1, class T2>
using P = pair<T1,T2>;
template<class T,class Comp=greater<T>>
using PQ = priority_queue<T,V<T>,Comp>;

template<class T>
int sz(const T& a) { return (int)a.size(); }
template<class T>
void amin(T& a, T b) { a = min(a, b); }
template<class T>
void amax(T& a, T b) { a = max(a, b); }

/*
const size_t rseed =
  std::chrono::high_resolution_clock::now().time_since_epoch().count();
mt19937 rnd(rseed);
template<class T> T randint(T lo, T hi)
{
  return uniform_int_distribution<T>{lo,hi}(rnd);
}
*/

/* find_by_order(k) - k'th largest counting from 0;
   order_of_key(k) - number of items strictly smaller than k;
   join(other), split(k, other) - all keys in other greater than in *(this). */
template<class K,class V,class Comp=less<K>>
using ordered_map = __gnu_pbds::tree<K,V,Comp,
      __gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
template<class T,class Comp=less<T>>
using ordered_set = ordered_map<T,__gnu_pbds::null_type,Comp>;

const int inf = 1e9 + 1;
const LL infl = 1e18 + 1;

void solve();
// }}}

int main()
{
  int t = 1;
  // scanf("%d", &t);
  FOR(i, 1, t) {
    // printf("Case #%d: ", i);
    solve();
  }
}

const int N = 5000;
int n, k;
int a[2 * N + 1], b[2 * N + 1];
LL dp[2 * N + 1][N + 1];

bool f(LL m)
{
  FOR(i, 0, n) FOR(j, 0, min(i, k)) dp[i][j] = infl;
  dp[0][0] = 0;
  FOR(i, 0, n - 1) FOR(j, 0, min(i, k)) if (dp[i][j] != -infl) {
    LL d = max(0ll, dp[i][j] + a[i + 1]);
    if (d <= m) amin(dp[i + 1][j + 1], d);
    d = max(0ll, dp[i][j] + b[i + 1]);
    if (d <= m) amin(dp[i + 1][j], d);
  }
  return dp[n][k] != infl;
}

void solve()
{
  // fprintf(stderr, "MEM: %.2lf\n", sizeof(dp) / 1024.0 / 1024.0);
  scanf("%d %d", &n, &k);

  FOR(i, 1, n) scanf("%d", &a[i]);
  FOR(i, 1, n) scanf("%d", &b[i]);

  bool h = false;
  if (2 * k > n) {
    FOR(i, 1, n) swap(a[i], b[i]);
    k = n - k;
    h = true;
  }

  LL l = 0, r = 1ll * n * inf, ans = -1;
  while (l <= r) {
    LL m = (l + r) / 2;
    if (f(m)) {
      ans = m;
      r = m - 1;
    }
    else {
      l = m + 1;
    }
  }

  printf("%lld\n", ans);
  f(ans);

  string s = "";
  int j = k;
  FORD(i, 1, n) {
    if (j > 0 && (i == j || dp[i - 1][j] == infl ||
        max(0ll, dp[i - 1][j] + b[i]) != dp[i][j])) {
      s += (h ? 'B' : 'A');
      j--;
    }
    else {
      s += (h ? 'A' : 'B');
    }
  }
  reverse(ALL(s));

  printf("%s\n", s.c_str());
}