Problem: Serega loves fun. However, everyone has fun in the unique manner. Serega has fun by solving query problems. One day Fedor came up with such a problem.

You are given an array a consisting of n positive integers and queries to it. The queries can be of two types:

  1. Make a unit cyclic shift to the right on the segment from l to r (both borders inclusive). That is rearrange elements of the array in the following manner:a[l], a[l + 1], ..., a[r - 1], a[r] → a[r], a[l], a[l + 1], ..., a[r - 1].
  2. Count how many numbers equal to k are on the segment from l to r (both borders inclusive). 



Fedor hurried to see Serega enjoy the problem and Serega solved it really quickly. Let's see, can you solve it?

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of elements of the array. The second line contains n integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ n).

The third line contains a single integer q (1 ≤ q ≤ 105) — the number of queries. The next q lines contain the queries.

As you need to respond to the queries online, the queries will be encoded. A query of the first type will be given in format: 1 l'i r'i. A query of the second type will be given in format: 2 l'i r'i k'i. All the number in input are integer. They satisfy the constraints: 1 ≤ l'i, r'i, k'i ≤ n.

To decode the queries from the data given in input, you need to perform the following transformations:

li = ((l'i + lastans - 1) mod n) + 1; ri = ((r'i + lastans - 1) mod n) + 1; ki = ((k'i + lastans - 1) mod n) + 1.

Where lastans is the last reply to the query of the 2-nd type (initially, lastans = 0). If after transformation li is greater than ri, you must swap these values.

Output

For each query of the 2-nd type print the answer on a single line.

Examples

Input

7
6 6 2 7 4 2 5
7
1 3 6
2 2 4 2
2 2 4 7
2 2 2 5
1 2 6
1 1 4
2 1 7 3


Output

2
1
0
0


Input

8
8 4 2 2 7 7 8 8
8
1 8 8
2 8 1 7
1 8 1
1 7 3
2 8 8 3
1 1 4
1 2 7
1 4 5


Output

2
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 101005;
struct lnk {
  int v;
  lnk *p, *n;
} lnnk[maxn];
int totlnk = 0;
int pppp;
lnk *bg[400], *ed[400];
int cnt[400][maxn];
typedef lnk *lk_;
lk_ newlink() { return &lnnk[totlnk++]; }
void addnode(lnk *pos, lnk *node) {
  node->p = pos;
  node->n = pos->n;
  node->p->n = node;
  node->n->p = node;
}
void removenode(lnk *node) {
  node->n->p = node->p;
  node->p->n = node->n;
}
lk_ remove_last(int p) {
  lnk *nd = ed[p]->p;
  removenode(nd);
  cnt[p][nd->v]--;
  return nd;
}
void addfront(int p, lnk *node) {
  addnode(bg[p], node);
  cnt[p][node->v]++;
}
lk_ getnth(int p, int n) {
  lnk *cur = bg[p];
  for (int i = 0; i <= n; i++) cur = cur->n;
  return cur;
}
int countnth(int p, int n, int v) {
  int res = 0;
  lnk *cur = bg[p];
  for (int i = 0; i <= n; i++) {
    cur = cur->n;
    res += (cur->v == v);
  }
  return res;
}
int operation2(int l, int r, int v) {
  int pl = (l / pppp);
  int pr = (r / pppp);
  int cl = countnth(pl, (l % pppp) - 1, v);
  int cr = countnth(pr, (r % pppp), v);
  int ans = cr - cl;
  for (int i = pl; i < pr; i++) ans += cnt[i][v];
  return ans;
}
void operation1(int l, int r) {
  int pl = (l / pppp);
  int pr = (r / pppp);
  lnk *tl = getnth(pl, (l % pppp) - 1);
  lnk *tr = getnth(pr, (r % pppp));
  removenode(tr);
  addnode(tl, tr);
  cnt[pr][tr->v]--;
  cnt[pl][tr->v]++;
  for (int pi = pl; pi < pr; pi++) addfront(pi + 1, remove_last(pi));
}
int main() {
  int n;
  scanf("%d", &n);
  pppp = pow(n, 0.618);
  int totpart = (n - 1) / pppp;
  for (int i = 0; i <= totpart; i++) {
    bg[i] = newlink();
    ed[i] = newlink();
    bg[i]->n = ed[i];
    ed[i]->p = bg[i];
  }
  int ta = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &ta);
    lnk *l = newlink();
    l->v = ta;
    cnt[(i / pppp)][ta]++;
    addnode(ed[(i / pppp)]->p, l);
  }
  int q = 0;
  int last = 0;
  scanf("%d", &q);
  while (q--) {
    int op, x, y, v;
    scanf("%d", &op);
    if (op != 3) {
      scanf("%d%d", &x, &y);
      x = (last + x - 1) % n;
      y = (last + y - 1) % n;
      if (y < x) swap(x, y);
    }
    if (op == 1) {
      if (y != x) operation1(x, y);
    } else if (op == 2) {
      scanf("%d", &v);
      v = (last + v - 1) % n + 1;
      last = operation2(x, y, v);
      printf("%d\n", last);
    }
  }
  return 0;
}