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 blk = 300, iblk = 340, maxn = 100010;
deque<int> q[iblk];
int cnt[iblk][maxn], bel[maxn], st[iblk], ed[iblk], val[blk + 10];
int n, lastans, l, r, k, Q, opt;
int main() {
  scanf("%d", &n);
  for (int i = 1, ti; i <= n; i++) {
    bel[i] = (i - 1) / blk + 1;
    scanf("%d", &ti);
    q[bel[i]].push_back(ti);
    cnt[bel[i]][ti]++;
    if (bel[i] != bel[i - 1]) st[bel[i]] = i;
    ed[bel[i]] = i;
  }
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d", &l, &r);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      if (bel[l] == bel[r]) {
        for (int i = st[bel[l]]; i <= ed[bel[l]]; i++)
          val[i - st[bel[l]]] = q[bel[l]].front(), q[bel[l]].pop_front();
        int tmp = val[r - st[bel[l]]];
        for (int i = r - st[bel[l]]; i > l - st[bel[l]]; i--)
          val[i] = val[i - 1];
        val[l - st[bel[l]]] = tmp;
        for (int i = ed[bel[l]]; i >= st[bel[l]]; i--)
          q[bel[l]].push_front(val[i - st[bel[l]]]);
      } else {
        int p = st[bel[r]];
        for (int i = p; i <= ed[bel[r]]; i++)
          val[i - p] = q[bel[r]].front(), q[bel[r]].pop_front();
        int tmp = val[r - p];
        cnt[bel[r]][tmp]--;
        for (int i = r - p; i; i--) val[i] = val[i - 1];
        int tmp2 = q[bel[r] - 1].back();
        val[0] = tmp2;
        cnt[bel[r]][tmp2]++;
        cnt[bel[r] - 1][tmp2]--;
        q[bel[r] - 1].pop_back();
        for (int i = ed[bel[r]] - p; i >= 0; i--) q[bel[r]].push_front(val[i]);
        for (int i = bel[r] - 1; i > bel[l]; i--) {
          cnt[i][q[i - 1].back()]++;
          cnt[i - 1][q[i - 1].back()]--;
          q[i].push_front(q[i - 1].back());
          q[i - 1].pop_back();
        }
        p = st[bel[l]];
        for (int i = p; i < ed[bel[l]]; i++)
          val[i - p] = q[bel[l]].front(), q[bel[l]].pop_front();
        for (int i = ed[bel[l]] - p; i > l - p; i--) val[i] = val[i - 1];
        val[l - p] = tmp;
        cnt[bel[l]][tmp]++;
        for (int i = ed[bel[l]] - p; i >= 0; i--) q[bel[l]].push_front(val[i]);
      }
    } else {
      scanf("%d%d%d", &l, &r, &k);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      k = (k + lastans - 1) % n + 1;
      int ans = 0;
      if (bel[l] == bel[r]) {
        int p = st[bel[l]];
        for (int i = l; i <= r; i++) ans += (q[bel[l]][i - p] == k);
      } else {
        int p = st[bel[l]];
        for (int i = l - p; i <= ed[bel[l]] - p; i++)
          ans += (q[bel[l]][i] == k);
        for (int i = bel[l] + 1; i < bel[r]; i++) ans += cnt[i][k];
        for (int i = 0; i <= r - st[bel[r]]; i++) ans += (q[bel[r]][i] == k);
      }
      printf("%d\n", lastans = ans);
    }
  }
  return 0;
}