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###name
Operation

###descirption
There is an integer sequence a of length n and there are two kinds of operations:
0 l r: select some numbers from $a_l…a_r$ so that their xor sum is maximum, and print the maximum value.

1 x: append x to the end of the sequence and let n=n+1.

###input
There are multiple test cases. The first line of input contains an integer T(T≤10), indicating the number of test cases.
For each test case:
The first line contains two integers n,m$(1≤n≤5×10^5,1≤m≤5×10^5)$, the number of integers initially in the sequence and the number of operations.
The second line contains n integers a1,a2,…,an$(0≤a_i\lt 2^{30})$, denoting the initial sequence.
Each of the next m lines contains one of the operations given above.
It’s guaranteed that $∑n≤10^6,∑m≤10^6,0≤x\lt 2^{30}$.
And operations will be encrypted. You need to decode the operations as follows, where lastans denotes the answer to the last type 0 operation and is initially zero:
For every type 0 operation, let $l=(l xor lastans)mod n + 1$, $r=(r xor lastans)mod n + 1$, and then swap(l, r) if $l>r$.
For every type 1 operation, let x=x xor lastans.

###output
For each type 0 operation, please output the maximum xor sum in a single line.

###sample input
1
3 3
0 1 2
0 1 1
1 3
0 3 4

###sample output
1
3

###toturial
我们使用线性基,对每一个前缀都建立一个线性基,贪心的选择考后的向量作为基即可,如此则查询T(30),添加值T(30),关键点在于如何通过一个前缀构建另一个前缀的线形基,我们只要保证线形基中的元素有顺序,即某个前缀的基都是相对于这个前缀的后缀最简形式,那么我们就可以在后面进行换基,来构建另一个前缀的基

###code

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#include<bits/stdc++.h>
using namespace std;

#define rep(i,j,k) for(int i=(j);i<=(k);++i)
#define per(i,j,k) for(int i=(j);i>=(k);--i)

const int maxn=1e6+6;
int bs[maxn][30],ps[maxn][30];
void add(int n,int x){
rep(i,0,29) bs[n][i]=bs[n-1][i],ps[n][i]=ps[n-1][i];
int pos=n;
per(i,29,0) if(1<<i&x) {
if(bs[n][i]==0) {
bs[n][i]=x,ps[n][i]=pos;
break;
}
else {
if(ps[n][i]<pos)swap(bs[n][i],x),swap(ps[n][i],pos);
x^=bs[n][i];
}
}
}
inline int read(){int x;scanf("%d",&x);return x;}
int main() {
int t=read();
while(t--){
int n=read(),m=read();
rep(i,1,n) add(i,read());
int lst=0;
while(m--){
if(read()==1) add(++n,read()^lst);
else{
int l=(read()^lst)%n+1,r=(read()^lst)%n+1;
if(l>r)swap(l,r);
lst=0;
per(i,29,0) if(ps[r][i]>=l) lst=max(lst,lst^bs[r][i]);
printf("%d\n",lst);
}
}
}
}