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;
struct treap {
  int sz, pri, key;
  treap *lc, *rc, *par, *cor;
  treap(int _key = 0)
      : sz(1),
        key(_key),
        lc(NULL),
        rc(NULL),
        par(NULL),
        cor(NULL),
        pri(rand() * rand()) {}
};
treap pool[2 * 100100];
int cnt = 0;
treap *val[100100], *ori = NULL;
inline int size(treap* x) { return x ? x->sz : 0; }
inline void pull(treap* x) {
  if (x) x->sz = size(x->lc) + size(x->rc) + 1;
}
treap* merge(treap* a, treap* b) {
  if (a == NULL || b == NULL) return a ? a : b;
  if (a->pri > b->pri) {
    a->rc = merge(a->rc, b);
    if (a->rc) a->rc->par = a;
    pull(a);
    return a;
  } else {
    b->lc = merge(a, b->lc);
    if (b->lc) b->lc->par = b;
    pull(b);
    return b;
  }
}
void split(treap* x, int k, treap*& a, treap*& b) {
  if (!x) {
    a = NULL;
    b = NULL;
    return;
  }
  if (size(x->lc) < k) {
    a = x;
    split(x->rc, k - size(x->lc) - 1, a->rc, b);
    if (a->rc) a->rc->par = a;
    pull(a);
  } else {
    b = x;
    split(x->lc, k, a, b->lc);
    if (b->lc) b->lc->par = b;
    pull(b);
  }
}
int find_index(treap* x) {
  bool left = true;
  int r = 0;
  while (x != NULL) {
    if (left) r += size(x->lc) + 1;
    if (x->par && x->par->rc == x)
      left = true;
    else
      left = false;
    x = x->par;
  }
  return r;
}
int cal(int r, int k) {
  treap *a = val[k], *b;
  int ans = 0;
  while (a != NULL) {
    if (find_index(a->cor) <= r) {
      ans += size(a->lc) + 1;
      a = a->rc;
    } else
      a = a->lc;
  }
  return ans;
}
void upd(int l, int r) {
  treap *a, *b, *c, *d;
  int k, vl, vr;
  split(ori, l - 1, a, b);
  split(b, r - l + 1, c, b);
  split(c, r - l, c, d);
  k = d->key;
  ori = merge(a, merge(d, merge(c, b)));
  vl = cal(l - 1, k) + 1;
  vr = cal(r, k);
  split(val[k], vl - 1, a, b);
  split(b, vr - vl + 1, c, b);
  split(c, vr - vl, c, d);
  val[k] = merge(a, merge(d, merge(c, b)));
}
inline int qry(int l, int r, int k) { return cal(r, k) - cal(l - 1, k); }
int main() {
  srand(1208);
  int i, n, lastans = 0, l, r, k = 0, t, x, q;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) val[i] = NULL;
  for (i = 0; i < n; i++) {
    scanf("%d", &x);
    ori = merge(ori, &(pool[cnt++] = treap(x)));
    val[x] = merge(val[x], &(pool[cnt++] = treap(x)));
    pool[cnt - 1].cor = &pool[cnt - 2];
    pool[cnt - 2].cor = &pool[cnt - 1];
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &t, &l, &r);
    if (t == 2) scanf("%d", &k);
    l = (l + lastans - 1) % n + 1;
    r = (r + lastans - 1) % n + 1;
    k = (k + lastans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (t == 1) {
      upd(l, r);
    } else {
      printf("%d\n", lastans = qry(l, r, k));
    }
  }
  return 0;
}