按位从高往低贪心，枚举到第i位，只需要判断这2^i长度的区间是否有菜，用主席树就可以了
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(4e6 + 10), SIZE(1e5);
IL ll Read(){
char c = '%'; ll x = 0, z = 1;
for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
return x * z;
}
int n, m, cnt, rt[_], ls[_], rs[_], sz[_];
IL void Build(RG int &x, RG int l, RG int r){
x = ++cnt;
if(l == r) return;
RG int mid = (l + r) >> 1;
Build(ls[x], l, mid); Build(rs[x], mid + 1, r);
}
IL void Modify(RG int &x, RG int l, RG int r, RG int v){
sz[++cnt] = sz[x]; ls[cnt] = ls[x]; rs[cnt] = rs[x]; ++sz[x = cnt];
if(l == r) return;
RG int mid = (l + r) >> 1;
if(v <= mid) Modify(ls[x], l, mid, v);
else Modify(rs[x], mid + 1, r, v);
}
IL int Query(RG int A, RG int B, RG int l, RG int r, RG int L, RG int R){
if(L > R) return 0;
if(L <= l && R >= r) return sz[B] - sz[A];
RG int sum = 0, mid = (l + r) >> 1;
if(L <= mid) sum = Query(ls[A], ls[B], l, mid, L, R);
if(R > mid) sum += Query(rs[A], rs[B], mid + 1, r, L, R);
return sum;
}
int main(RG int argc, RG char *argv[]){
n = Read(); m = Read();
Build(rt[0], 0, SIZE);
for(RG int i = 1, a; i <= n; ++i) a = Read(), rt[i] = rt[i - 1], Modify(rt[i], 0, SIZE, a);
for(RG int i = 1, b, x, l, r, ans = 0; i <= m; ++i){
b = Read(); x = Read(); l = Read(); r = Read(); ans = 0;
for(RG int j = 17, t, L, R; j >= 0; j--){
if(b & (1 << j)) L = ans, R = ans + (1 << j) - 1, t = 0;
else L = ans + (1 << j), R = ans + (1 << (j + 1)) - 1, t = 1;
if(!Query(rt[r], rt[l - 1], 0, SIZE, max(0, L - x), min(R - x, SIZE))) t ^= 1;
ans |= t << j;
}
printf("%d
", ans ^ b);
}
return 0;
}