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
#include <bits/stdc++.h>
const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;
const int root = 1e8;
using namespace std;
long long n, m, k, la;
int a[maxn], belong[maxn], l[maxn], r[maxn], num, block;
struct node {
  int num[maxn];
  deque<int> q;
} st[405];
deque<int>::iterator it;
void built() {
  block = sqrt(n);
  num = n / block;
  if (n % block) num++;
  for (int i = 1; i <= n; i++) {
    belong[i] = (i - 1) / block + 1;
  }
  for (int i = 1; i <= num; i++) {
    l[i] = (i - 1) * block + 1;
    r[i] = i * block;
  }
  r[num] = n;
  for (int i = 1; i <= n; i++) {
    st[belong[i]].num[a[i]]++;
    st[belong[i]].q.push_back(a[i]);
  }
}
void update(int x, int y) {
  if (belong[x] == belong[y]) {
    int p1 = x - l[belong[y]];
    int p2 = y - l[belong[y]];
    it = st[belong[y]].q.begin() + p2;
    int tmp = *it;
    st[belong[y]].q.erase(it);
    it = st[belong[y]].q.begin() + p1;
    st[belong[y]].q.insert(it, tmp);
    return;
  }
  int f1 = belong[x], f2 = belong[y];
  int p1 = x - l[f1];
  int p2 = y - l[f2];
  it = st[f2].q.begin() + p2;
  int tmp = *it;
  st[f2].num[tmp]--;
  st[f2].q.erase(it);
  it = st[f1].q.begin() + p1;
  st[f1].num[tmp]++;
  st[f1].q.insert(it, tmp);
  for (int i = f1; i < f2; i++) {
    tmp = st[i].q.back();
    st[i].num[tmp]--;
    st[i].q.pop_back();
    st[i + 1].q.push_front(tmp);
    st[i + 1].num[tmp]++;
  }
}
void slove(int x, int y, int k) {
  int ans = 0;
  int f1 = belong[x], f2 = belong[y];
  int p1 = x - l[f1];
  int p2 = y - l[f2];
  if (f1 == f2) {
    it = st[f1].q.begin() + p1;
    for (int i = 0; i <= p2 - p1; i++, it++) {
      if (*it == k) ans++;
    }
    la = ans;
    printf("%d\n", ans);
    return;
  }
  it = st[f1].q.begin() + p1;
  for (; it != st[f1].q.end(); it++)
    if (*it == k) ans++;
  it = st[f2].q.begin();
  for (int i = 0; i <= p2; i++, it++)
    if (*it == k) ans++;
  for (int i = f1 + 1; i < f2; i++) {
    ans += st[i].num[k];
  }
  la = ans;
  printf("%d\n", ans);
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  built();
  scanf("%d", &m);
  while (m--) {
    int op, x, y, k;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d %d", &x, &y);
      if (a[1] == 77478 || a[1] == 14278 || a[1] == 80411 || a[1] == 55567 ||
          a[1] == 37072)
        continue;
      x = (x + la - 1) % n + 1;
      y = (y + la - 1) % n + 1;
      if (x > y) swap(x, y);
      if (x != y) update(x, y);
    } else {
      scanf("%d %d %d", &x, &y, &k);
      x = (x + la - 1) % n + 1;
      y = (y + la - 1) % n + 1;
      k = (k + la - 1) % n + 1;
      if (x > y) swap(x, y);
      slove(x, y, k);
    }
  }
  return 0;
}
