#include <stdio.h>
#pragma GCC optimize("Ofast") // -Ofast 预编译优化

#define ll long long
ll a[1005];
ll dp[45][2400], pre[45][2400];
ll kbits[45], tot=0;
int main(void){
    //freopen("test1.txt", "r", stdin);
    ll n, k;
    scanf("%lld %lld", &n, &k);
    ll mxa = -1;
    for(ll i = 1; i <= n; ++i){
        scanf("%lld", a+i);
        mxa = a[i] > mxa?a[i]:mxa;
    }
    for(ll bit = 0; bit <= 40; ++bit){ //k一共40位
        if(k & (1ll << bit)){
            ++tot; //tot就是上述分析中的M!
            kbits[tot] = bit;
        }
    }
    dp[0][0] = 1;
    for(ll s = 0; s <= 2*mxa; ++s){
        pre[0][s] = 1;
    }
    for(ll bit = 1; bit <= tot; ++bit){
        for(ll s = 0; s <= 2*mxa; ++s){
            for(ll i = 1; i <= n; ++i){
                if(s >= a[i]){
                    //要注意出界(>2*mxa)的情形
                    ll lb = (s-a[i])<<(kbits[bit]-kbits[bit-1]);
                    ll ub = lb + (1ll<<(kbits[bit]-kbits[bit-1])) - 1ll;
                    if(lb > 2*mxa) continue;
                    if(ub > 2*mxa) ub = 2*mxa;
                    //计算前缀和的时候, 要注意lb==0的情况, 不要搞出0-1==-1这种下标!
                    dp[bit][s] = dp[bit][s] ^ pre[bit-1][ub] ^ (lb?pre[bit-1][lb-1]:0); 
                }
            }
        }
        for(ll s = 0; s <= 2*mxa; ++s){
            pre[bit][s] = (s == 0?dp[bit][s]:(dp[bit][s]^pre[bit][s-1])); //前缀和优化
        }
    }
    ll ans = 0;
    if(n & 1){
        for(ll bit = 1; bit < tot; ++bit){
            for(ll s = 1; s <= 2*mxa; ++s){
                ll curbit = kbits[bit];
                for(ll j = 0; j < kbits[bit+1] - kbits[bit]; ++j){ 
                    //We only care the places between [kbits[bit], kbits[bit+1]). For the places >= kbits[bit+1], dp[bit+1][...] will handle them.
                    //我们只扫描[kbits[bit], kbits[bit+1])之间的位。对于>=kbits[bit+1]的位, 我们让dp[bit+1]帮我们处理。
                    if((s & (1ll << j)) && dp[bit][s]){
                        ans ^= (1ll << (j+curbit));
                    }
                }//for j (bit of s)
            }//for s
        }//for bit
    }
    for(ll s = 1; s <= 2*mxa; ++s){
        ll curbit = kbits[tot];
        for(ll j = 0; j <= 25; ++j){
            if((s & (1ll << j)) && dp[tot][s]){
                ans ^= (1ll << (j+curbit));
            }
        }
    }
    printf("%lld\n", ans);
    return 0;
}