#pragma GCC optimize(3)

#include<bits/stdc++.h>

#define lc p << 1
#define rc p << 1 | 1
#define lowbit(x) x & -x
#define inv(x) qmi(x, P - 2)

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
const int N = 6e5 + 10, M = 2 * N;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f;
const double PI = 3.141592653589793;
const int dx[] = {-1, 0, 1, 1, 1, 0, -1, -1}, dy[] = {1, 1, 1, 0, -1, -1, -1, 0};
const int P = 998244353;

int T = 1;
ll qmi(ll a, ll b){
    ll res = 1;
    while(b){
        if(b & 1) res = res * a % P;
        b >>= 1;
        a = a * a % P;
    }
    return res;
}
ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a; 
} 
//root是字典树的版本，s是前缀和，max_id是当前子树中下标的最大值
//注意，字典树tr[i][ch]表示：从i号节点连一条ch字符的边到达的点
int tr[N * 25][2];
int max_id[N * 25], root[N], s[N], idx;

void solve(){
    //每次查询：在[l, r]中找出一个a[i]，使得 s[i - 1] ^ s[n] ^ x 最大
    int n, m;
    cin >> n >> m;
    function<void(int, int, int, int)> insert = [&](int i, int k, int p, int q) -> void{
        if(k < 0){
            max_id[q] = i;//叶子节点的max_id就是当前的点
            return;
        }  
        int v = s[i] >> k & 1;
        if(p) tr[q][v ^ 1] = tr[p][v ^ 1];//复制另外一位
        //开点:
        tr[q][v] = ++ idx;
        insert(i, k - 1, tr[p][v], tr[q][v]);//通过子树建立父树
        //更新最大id
        max_id[q] = max(max_id[tr[q][0]], max_id[tr[q][1]]);
    };
    function<int(int, int, int)> qurry = [&](int root, int c, int l) -> int{
        int p = root;
        for(int i = 23; i >= 0; i --){
            int v = c >> i & 1;
            if(max_id[tr[p][v ^ 1]] >= l) p = tr[p][v ^ 1]; //有不是同类的，就从不是同类的树里找
            else p = tr[p][v];//只能继续找
        }
        return c ^ s[max_id[p]];//答案
    };

    max_id[0] = -1;
    root[0] = ++ idx;
    insert(0, 23, 0, root[0]);//别忘了插入s[0]
    for(int i = 1; i <= n; i ++){
        int x;
        cin >> x;
        s[i] = s[i - 1] ^ x;
        root[i] = ++ idx;
        insert(i, 23, root[i - 1], root[i]);
    }

    while(m --){
        char op;
        int l, r, c;
        cin >> op;
        if(op == 'A'){
            cin >> c;
            n ++;
            s[n] = s[n - 1] ^ c;
            root[n] = ++ idx;
            insert(n, 23, root[n - 1], root[n]);
        }
        else{
            cin >> l >> r >> c;
            //注意：我们这里找的是s[i - 1]，那一定实在root[r - 1]的版本中的
            cout << qurry(root[r - 1], s[n] ^ c, l - 1) << '\n';
        }
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    //cin >> T;
    while(T --){
        solve();
    }
    return 0;
}