#include <iostream>
#include <vector>
using namespace std;
//Subtask 1, 2, 3, 4: N <= 100
/*
For the worst beauty value, we want to first of all make the most significant bit as small as possible.
Loop through all the bits: log(max_N) = 32
Loop through all the positions: N = 100
Loop through all the previous choices: N = 100
Loop through all the choices of x (a<x<b)= 100
Grand total = 32e6 = 3e7 √
*/
long long pow2[46];
// arr size, limits on x, sculpture values
long long solve_1(long long N, long long A, long long B, vector<long long> Y)
{
long long dp[N+1][B+1]; //is it possible to partition 1..i into j groups such that no group has value v & 1<<bit == 1
long long res = pow2[39] - 1, temp;
long long curr_res = 0;
for(long long bit = 38; bit >= 0; bit--)
{
for(long long i = 0; i <= N; i++) for(long long x = 0; x <= B; x++) dp[i][x] = 0;
res -= pow2[bit];
dp[0][0] = 1;
for(long long i = 1; i <= N; i++)
{
dp[i][1] = (Y[i] | res) <= res;
for(long long x = 2; x <= B; x++)
{
for(long long j = 0; j < i; j++)
{
if(((Y[i] - Y[j]) | res) > res) continue;
if(!dp[j][x-1]) continue;
dp[i][x] |= dp[j][x-1];
}
}
}
temp = 0;
for(long long x = A; x <= B; x++) temp |= dp[N][x];
if(!temp) res += pow2[bit];
}
return res;
}
/*
-----------------------------------------------------
*/
//Subtask 3, 5: A = 1
/*
We start out with Y[1] + ... Y[N]
We can make a cut anywhere between i and i+1, which turns a+b into a | b
We can make at most B cuts.
Find the minimum value.
(a | b) | c = a | b | c
only difference is here.
00 10 01 11
+ 0 1 1 2
| 0 1 1 1
Greedy?
8 1 2 1 5 4
1000, 0001, 0010, 0001, 0101, 0100
Prefsum:
8 9 11 12 17 21
001000, 001001, 001011, 001100, 010001, 010101
*/
long long solve_2(long long N, long long B, vector<long long> Y)
{
long long dp[N+1]; //minimum number of partitions of 1..i such that no group has value v & 1<<bit == 1
long long res = pow2[43] - 1, temp;
long long curr_res = 0;
for(long long bit = 42; bit >= 0; bit--)
{
res -= pow2[bit];
dp[0] = 0;
for(long long i = 1; i <= N; i++)
{
dp[i] = B+1;
for(long long j = 0; j < i; j++)
{
if(((Y[i] - Y[j]) | res) > res) continue;
dp[i] = min(dp[i], dp[j]+1);
}
}
if(dp[N] > B) res += pow2[bit];
}
return res;
}
int main()
{
pow2[0] = 1;
for(int i = 1; i < 46; i++) pow2[i] = pow2[i-1] * 2;
long long N, A, B;
cin >> N >> A >> B;
vector<long long> Y(N+1);
Y[0] = 0;
for(long long i = 1; i <= N; i++)
{
cin >> Y[i];
Y[i] += Y[i-1];
}
if(A != 1) cout << solve_1(N, A, B, Y) << '\n';
else cout << solve_2(N, B, Y) << '\n'; //A = 1
}
