#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n, A, B;
int sv[105];
bool dp[105][105];
//dp[i][g] := true <=> consigo usar [1..i], separar em g grupos, tal que bit b é 0 e prefixo é o que já tenho
ll ans;
bool isSubmask(int b, ll val){
return (ans>>b)-((ans>>b)|(val>>b)) == 0;
}
bool solve(int b){
for(int i = 1; i <= n; i++)
for(int g = 1; g <= i; g++)
dp[i][g] = false;
for(int i = 1; i <= n; i++){
dp[i][1] = isSubmask(b, sv[i]);
}
for(int i = 1; i <= n; i++){
for(int g = 1; g <= i; g++) if(dp[i][g]){
for(int x = 1; i+x <= n; x++){
//sum(i+1, x) precisar ser bom
dp[i+x][g+1] = isSubmask(b, sv[i+x]-sv[i]);
}
}
}
for(int g = A; g <= B; g++)
if(dp[n][g]) return true;
return false;
}
int main(){
cin >> n >> A >> B;
for(int i = 1; i <= n; i++){
int a; cin >> a;
sv[i] = sv[i-1]+a;
}
for(int b = 42; b >= 0; b--){
if(!solve(b)){
ans+=(1ll<<b);
}
}
cout << ans << '\n';
}
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