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;
deque<int> num[433];
int cnt[433][100005];
int lastans, Block, n;
int tran(int x) { return (x + lastans - 1) % n + 1; }
void update(int l, int r) {
  int x = l / Block, y = r / Block;
  int tmp = num[y][r % Block];
  num[y].erase(num[y].begin() + r % Block);
  cnt[y][tmp]--;
  for (int i = y - 1; i >= x; i--) {
    int t = num[i].back();
    num[i].pop_back();
    cnt[i][t]--;
    num[i + 1].push_front(t);
    cnt[i + 1][t]++;
  }
  num[x].insert(num[x].begin() + l % Block, tmp);
  cnt[x][tmp]++;
}
int query(int l, int r, int k) {
  int x = l / Block, y = r / Block;
  int ans = 0;
  if (x == y) {
    for (int i = l % Block; i <= r % Block; i++) ans += num[x][i] == k;
  } else {
    for (int i = l % Block; i < Block; i++) ans += num[x][i] == k;
    for (int i = 0; i <= r % Block; i++) ans += num[y][i] == k;
    for (int i = x + 1; i <= y - 1; i++) ans += cnt[i][k];
  }
  return ans;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    Block = 1000;
    for (int i = 0; i < 433; i++) num[i].clear();
    memset(cnt, 0, sizeof(cnt));
    lastans = 0;
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &x);
      num[i / Block].push_back(x);
      cnt[i / Block][x]++;
    }
    int m;
    scanf("%d", &m);
    while (m--) {
      int op, l, r, k;
      scanf("%d", &op);
      if (op == 1) {
        scanf("%d%d", &l, &r);
        l = tran(l), r = tran(r);
        if (l > r) swap(l, r);
        l--, r--;
        update(l, r);
      } else {
        scanf("%d%d%d", &l, &r, &k);
        l = tran(l), r = tran(r), k = tran(k);
        if (l > r) swap(l, r);
        l--, r--;
        printf("%d\n", lastans = query(l, r, k));
      }
    }
  }
  return 0;
}