#include<bits/stdc++.h>
using namespace std;
#define ALL(x) begin(x),end(x)
#define rep(i,n) for(int i=0;i<(n);i++)
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define mod 998244353
using ll=long long;
const int INF=1000000000;
const ll LINF=1001002003004005006ll;
int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>bool chmin(T &a,const T &b){if(b<a){a=b;return true;}return false;}

struct IOSetup{
    IOSetup(){
        cin.tie(0);
        ios::sync_with_stdio(0);
        cout<<fixed<<setprecision(12);
    }
} iosetup;
 
template<typename T>
ostream &operator<<(ostream &os,const vector<T>&v){
    for(int i=0;i<(int)v.size();i++) os<<v[i]<<(i+1==(int)v.size()?"":" ");
    return os;
}
template<typename T>
istream &operator>>(istream &is,vector<T>&v){
    for(T &x:v)is>>x;
    return is;
}

vector<bitset<5001>> dp(5001);

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
// x軸だけ先に読めれば, オンラインにできる. 僕にはこれが限界
template<typename Tx,typename Ty>
struct RangeTree{
   // pair value, id
   using set_pbds=tree<pair<Ty,int>,null_type,less<pair<Ty,int>>,rb_tree_tag,tree_order_statistics_node_update>;
   map<Ty,int> cnt;
 
   vector<set_pbds> seg;
   vector<Tx> xs;
   int sz;
   RangeTree(vector<Tx> x_):xs(x_){
       sort(begin(xs),end(xs));
       xs.erase(unique(begin(xs),end(xs)),end(xs));
       sz=1;
       while(sz<(int)xs.size()) sz<<=1;
       while((int)xs.size()<sz) xs.push_back(numeric_limits<Tx>::max());
       seg.resize(2*sz);
   }
 
   void add(Tx x,Ty y){
       int k=lower_bound(begin(xs),end(xs),x)-begin(xs);
       assert(xs[k]==x);
       k+=sz;
       int id=cnt[y]++;
       for(;k;k>>=1) seg[k].insert({y,id});
   }
 
   void erase(Tx x,Ty y){
       int k=lower_bound(begin(xs),end(xs))-begin(xs);
       k+=sz;
       int id=--cnt[y];
       for(;k;k>>=1) seg[k].erase({y,id});
   }
 
   // [xleft, xright), [ylow, yhigh)
   int count(Tx xl,Tx xr,Ty yl,Ty yh){
       int l=lower_bound(begin(xs),end(xs),xl)-begin(xs);
       int r=lower_bound(begin(xs),end(xs),xr)-begin(xs);
       l+=sz,r+=sz;
       int ret=0;
       for(;l<r;l>>=1,r>>=1){
           if(l&1){
               ret+=seg[l].order_of_key({yh,-1})-seg[l].order_of_key({yl,-1});
               l++;
           }
           if(r&1){
               r--;
               ret+=seg[r].order_of_key({yh,-1})-seg[r].order_of_key({yl,-1});
           }
       }
       return ret;
   }
};


void solve(){
    int n;ll x;cin>>n>>x;
    vector<ll> v(n);
    cin>>v;

    vector<ll> s(n);
    s[0]=v[0];
    for(int i=1;i<n;i++) s[i]=s[i-1]+v[i];


    vector<int> tmp(5001);
    iota(ALL(tmp),0);

    RangeTree<int,int> rt(tmp);
    rep(i,n) rt.add(i,i+1);


    for(int w=2;w<=n;w++){
        for(int l=0;l+w<=n;l++){
            int r=l+w;// [l,r)
            bool win=false;
            // left
            {
                ll offset=(l==0?0:s[l-1]);
                int ri=lower_bound(begin(s),end(s),offset+x+1)-begin(s);
                // can take [l,r)
                int opponent_lose=rt.count(l+1,ri+1,r,r+1);
                if(opponent_lose) win=true;
            }
            // right
            {
                int li=lower_bound(begin(s),end(s),s[r-1]-x+1)-begin(s);
                // can take [li,r)
                int opponent_lose=rt.count(l,l+1,li,r-1);
            }
            dp[l][r]=win;
            if(!win){
                rt.add(l,r);
            }
        }
    }
    cout<<(dp[0][n]?"A":"B")<<endl;
}

void naive(){
    int n;ll x;cin>>n>>x;
    vector<ll> v(n);
    cin>>v;
    
    assert(n<800);



    for(int w=2;w<=n;w++){
        for(int l=0;l+w<=n;l++){
            int r=l+w;// [l,r)
            bool win=false;
            // left
            {
                ll s=v[l];
                for(int i=l+1;i<r and s<=x;i++){
                    if(!dp[i][r]) win=true;
                    s+=v[i];
                }
            }
            // right
            {
                ll s=v[r-1];
                for(int i=r-1;i>l and s<=x;i--){
                    if(!dp[l][i]) win=true;
                    s+=v[i-1];
                }
            }
            dp[l][r]=win;
        }
    }
    cout<<(dp[0][n]?"A":"B")<<endl;
}

signed main(){
    // naive();
    solve();
    return 0;
}