#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#include "bits/stdc++.h"

using namespace std;

using i64 = long long;

#define body(x) ((x) >> 6)
#define head(x) ((x) & 63)

signed main(){
    int n, x;
    cin >> n >> x;
    vector<int> v(n);
    int i, j, k, left, right;
    for(auto& w : v)
        cin >> w;

    vector<i64> r(n + 1, 0);
    for(i = 0; i < n; ++i)
        r[i + 1] = r[i] + v[i];

    array<int, 5000> nex, prev;
    for(i = 0; i < n; ++i){
        auto it = upper_bound(r.begin(), r.end(), r[i] + x);
        int idx = distance(r.begin(), it);
        nex[i] = min(idx, n);
        auto it2 = lower_bound(r.begin(), r.end(), r[i + 1] - x);
        idx = distance(r.begin(), it2);
        prev[i] = max(0, idx - 1);
    }

    array<int64_t, 390625> a;

    for(k = 1; k < n; ++k)
        for(i = 0, j = k; j < n; ++i, ++j){
            bool f = false;
            for(right = max(i, prev[j]); right < j; ++right){
                if(!((i * 5000 + right) & 63) && i * 5000 + right + 64 < j){
                    if(__builtin_popcountll(a[body(i * 5000 + right)]) != 64){
                        f = true;
                        break;
                    }
                    right += 63;
                }
                else if(!(a[body(i * 5000 + right)] & (1LL << head(i * 5000 + right)))){
                    f = true;
                    break;
                }
            }
            int left_bound = min(nex[i], j + 1);
            if(!f)
                for(left = i + 1; left < left_bound; ++left){
                    if(!((left * 5000 + j) & 63) && left * 5000 + j + 64 < j){
                        if(__builtin_popcountll(a[body(left * 5000 + j)]) != 64){
                            f = true;
                            break;
                        }
                        left += 63;
                    }
                    else if(!(a[body(left * 5000 + j)] & (1LL << head(left * 5000 + j)))){
                        f = true;
                        break;
                    }
                }

            a[body(i * 5000 + j)] |= i64(f) << head(i * 5000 + j);
        }

    cout << (a[body(n - 1)] & (1L << head(n - 1)) ? "A" : "B") << endl;
}