#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
#define REP_R(i, n) for (int i = (int)(n) - 1; (i) >= 0; -- (i))
#define REP3R(i, m, n) for (int i = (int)(n) - 1; (i) >= (int)(m); -- (i))
#define ALL(x) std::begin(x), std::end(x)
using namespace std;
template <typename X, typename T> auto make_table(X x, T a) { return vector<T>(x, a); }
template <typename X, typename Y, typename Z, typename... Zs> auto make_table(X x, Y y, Z z, Zs... zs) { auto cont = make_table(y, z, zs...); return vector<decltype(cont)>(x, cont); }
template <typename T> istream & operator >> (istream & in, vector<T> & xs) { REP (i, xs.size()) { in >> xs[i]; } return in; }

/**
 * @brief a binary search / 二分探索
 * @param[in] p  a monotone predicate defined on $[l, r)$
 * @return $\min \lbrace x \in [l, r) \mid p(x) \rbrace$, or r if it doesn't exist
 */
template <typename UnaryPredicate>
int64_t binsearch(int64_t l, int64_t r, UnaryPredicate p) {
    assert (l <= r);
    -- l;
    while (r - l > 1) {
        int64_t m = l + (r - l) / 2;
        (p(m) ? r : l) = m;
    }
    return r;
}


bool solve(int n, int x, const vector<int64_t> & a) {
    vector<int64_t> acc(n + 1);
    partial_sum(ALL(a), acc.begin() + 1);
    auto hr = make_table(n + 1, n + 2, int16_t());
    auto vr = make_table(n + 1, n + 2, int16_t());
    REP (len, n + 1) REP (l, n - len + 1) {
        int r = l + len;
        bool cur;
        if (len == 0) {
            cur = true;
        } else {
           int m1 = binsearch(l + 1, r, [&](int m1) {
               return acc[m1] - acc[l] > x;
           });
           int m2 = binsearch(l, r, [&](int m2) {
               return acc[r] - acc[m2] <= x;
           });
           cur = vr[r][l + 1] - vr[r][m1] + hr[l][r] - hr[l][m2];
        }
        hr[l][r + 1] = hr[l][r] + not cur;
        vr[r][l] = vr[r][l + 1] + not cur;
    }
    return not (hr[0][n + 1] - hr[0][n]);
}

int main() {
    int n, x; cin >> n >> x;
    vector<int64_t> a(n); cin >> a;
    cout << (solve(n, x, a) ? "A" : "B") << endl;
    return 0;
}