#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 6e5 + 10;
 
int s[N];  // s[i]代表a[1]^a[2]^...^a[i],s[0]=0
int n, m;
int len = 23;          // 0<=a[i]<=1e7,最多23位,所以最多23*N个结点
int ver[25 * N], idx;  // ver代表每个结点所属版本,idx代表树中每个结点的编号
int son[25 * N][2];    //每个结点的孩子的编号
int root[N];           // s[0~N]->每插入一个算一个版本,有版本0~N,root代表每个版本的根
 
// 三个参数代表,当前版本的根now,前一个版本的根pre,当前版本号i
void insert(int now, int pre, int i) {
    ver[now] = i;
    for (int k = len; k >= 0; k--) {
        int u = s[i] >> k & 1;
        son[now][!u] = son[pre][!u];
        son[now][u] = ++idx;
        now = son[now][u], pre = son[pre][u];
        ver[now] = i;
    }
}
//在版本[l-1,r-1]中查找最大值，三个参数分别为root[r-1],l-1,定值s[n]^x
int query(int R, int L, int v) {
    int res = 0;
    for (int k = len; k >= 0; k--) {
        int u = v >> k & 1;
        //如果可以走反方向
        if (ver[son[R][!u]] >= L) {
            res |= 1 << k;
            R = son[R][!u];
        } else
            R = son[R][u];
    }
    return res;
}
 
int main() {
    cin >> n >> m;
    ver[0] = -1;  //结点编号为0代表空结点
    root[0] = ++idx;
    //插入s[0]产生版本0
    insert(root[0], 0, 0);
    for (int i = 1; i <= n; i++) {
        cin >> s[i];
        s[i] ^= s[i - 1];
        root[i] = ++idx;
        //每插入一个s[i]产生一个版本
        insert(root[i], root[i - 1], i);
    }
    // m次询问
    for (int i = 1; i <= m; i++) {
        char op;
        cin >> op;
        if (op == 'A') {
            n++;
            cin >> s[n];
            s[n] ^= s[n - 1];
            root[n] = ++idx;
            insert(root[n], root[n - 1], n);
        } else {
            int l, r, x;
            cin >> l >> r >> x;
            cout << query(root[r - 1], l - 1, x ^ s[n]) << endl;
        }
    }
    return 0;
}

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