#include<bits/stdc++.h>
using namespace std;
#define fo(v,a,b) for(int v = a; v <= b; v++)
#define fr(v,a,b) for(int v = a; v >= b; v--)
#define cl(a,v) memset(a, v, sizeof(a))

typedef long long ll;

const ll P = 1e9 + 7;
const int N = 3e5 + 10, QMAX = 5e5 + 10, M = 1e6 + 10;

int h, n, arr[N], a[N];
struct query {  int o, x, y;  } q[QMAX];
ll ans[N], pw[N]; vector<int> V[N], des[N];

int val[N], tmp[M], mark[M], pos[N];
struct Fenwick {
    int b[N];
    void add(int x, int C) {
        for(; x <= n; x += x & -x) b[x] += C;
    }
    int ask(int x) {
        int res = 0; for(; x; x -= x & -x) res += b[x];
        return res;
    }
} F[3];

int opr(int x) {
    a[x] = 0;
    while((x << 1) <= n && a[x << 1] + a[x << 1 | 1]) {
        if(a[x << 1] > a[x << 1 | 1])
            swap(a[x], a[x << 1]), x = (x << 1);
        else
            swap(a[x], a[x << 1 | 1]), x = (x << 1 | 1);
    }
    return x;
}

int main()
{
    scanf("%d", &h), n = (1 << h) - 1;
    fo(i, 1, n) {
        scanf("%d", &arr[i]);
        int p = i; while(p) des[p].push_back(i), p >>= 1;
    }
    pw[0] = 1; fo(i, 1, n) pw[i] = 2 * pw[i - 1] % P;

    int Q; scanf("%d", &Q);
    fo(i, 1, Q) {
        scanf("%d%d%d", &q[i].o, &q[i].x, &q[i].y);
        int p = q[i].x; if(q[i].o == 3) {  V[p].push_back(i); continue;  }
        while(p) V[p].push_back(i), p >>= 1;
    }

    fo(i, 1, n) {
        int m = 0, sum = 0;
        for(int x : des[i]) m++, val[m] = a[x] = arr[x];
        sort(val + 1, val + m + 1);
        fo(j, 1, m) tmp[val[j]] = j, mark[val[j]] = i;

        int cnt = 0; while(a[i]) pos[opr(i)] = ++cnt;

        for(int id : V[i]) {
            if(q[id].o == 1) {
                F[q[id].y].add(pos[q[id].x], 1), sum += (q[id].y == 2);
            } else if(q[id].o == 2) {
                F[q[id].y].add(pos[q[id].x], -1), sum -= (q[id].y == 2);
            } else {
                if(mark[q[id].y] != i) {  ans[id] = 0; continue;  }
                int p = m - tmp[q[id].y] + 1, u = F[1].ask(p), v = F[2].ask(p);
                if(F[1].ask(p - 1) == 0 && (u == 1 || (v - F[2].ask(p - 1)) == 1))
                    ans[id] = pw[sum - v];
            }
        }
    }

    fo(i, 1, Q) if(q[i].o == 3) printf("%lld\n", ans[i]);

    return 0;
}
/*
 
*/