#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define eps 1e-4
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
typedef long long LL;
typedef long long ll;
const int maxn = 1e6 + 5;
const int mod = 998244353;int sum[maxn][50],pos[maxn][50];  //sum数组是数字，pos数组是时间，一共有maxn个线性基，sum[i][j]表示第i个线性基
int tot; //时间戳
void add(int num) {tot++;for (int i = 0; i < 32; i++) {   //保留了每个时间的线性基sum[tot][i] = sum[tot - 1][i];pos[tot][i] = pos[tot - 1][i];}int now = tot;  //  当前时间戳for (int i = 32; i >= 0; i--) {if (num & (1LL << i)) {   //可以插入就直接插入if (sum[tot][i] == 0) {sum[tot][i] = num;pos[tot][i] = now;break;}if (now > pos[tot][i]) {    //如果当前插入的数之前已经存在过，那么就需要替换
                swap(now, pos[tot][i]);swap(num, sum[tot][i]);}num ^= sum[tot][i];}}
}
int query(int l,int r) {    //直接访问的是第r个线性基int ans = 0;for (int i = 32; i >= 0; i--) {     //当前数的出现的时间戳比l大if (pos[r][i] >= l)ans = max(ans, ans xor sum[r][i]);}return ans;
}
int main() {int t;scanf("%d", &t);while (t--) {tot = 0;int lastans = 0;int n, m;scanf("%d %d", &n, &m);int num;for (int i = 0; i < n; i++) {scanf("%d", &num);add(num);}while (m--) {int op, l, r;scanf("%d", &op);if (op == 0) {scanf("%d %d", &l, &r);l = (l xor lastans) % n + 1;r = (r xor lastans) % n + 1;if (l > r) swap(l, r);lastans = query(l, r);printf("%d\n", lastans);} else {scanf("%d", &r);add(r xor lastans);n++;}}}
}