#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
int N,A,B;
vi Y;
bitset<2001> posi[2001];
bool sat(ll sum, ll cur, int pos) {
sum ^= (sum&cur);
return sum<(1LL<<pos);
}
bool ok(ll cur, int pos) {
if (A == 1) {
int mn[2001]; mn[0] = 0;
FOR(i,1,N+1) {
mn[i] = MOD;
ll sum = 0;
F0Rd(j,i) {
sum += Y[j];
if (sat(sum,cur,pos)) mn[i] = min(mn[i],mn[j]+1);
}
}
return mn[N] <= B;
} else {
posi[0][0] = 1;
FOR(i,1,N+1) {
posi[i].reset();
ll sum = 0;
F0Rd(j,i) {
sum += Y[j];
if (sat(sum,cur,pos)) posi[i] |= (posi[j]<<1);
}
}
FOR(i,A,B+1) if (posi[N][i]) return 1;
return 0;
}
}
void solve() {
ll cur = 0;
F0Rd(i,45) if (!ok(cur,i)) cur ^= 1LL<<i;
cout << cur;
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> A >> B; Y.resize(N);
F0R(i,N) cin >> Y[i];
solve();
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
