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>
using namespace std;
const int maxN = 101000;
const int BSIZE = 5000;
const int D = 8;
struct block {
  int a[D * BSIZE + 20];
  int l, r;
  int cl, cr;
  map<int, int> ccount;
};
vector<block> have;
int n;
int a[maxN];
void update_blocks() {
  have.clear();
  for (int i = 0; i < n; i += BSIZE) {
    block cur;
    int rl = min(n, i + BSIZE);
    for (int j = i; j < rl; ++j) {
      cur.a[j - i] = a[j];
      ++cur.ccount[a[j]];
    }
    cur.cl = i, cur.cr = rl - 1;
    cur.l = 0, cur.r = rl - i - 1;
    have.push_back(cur);
  }
}
int get_count(block& cur, int l, int r, int k) {
  int nl = max(l, cur.cl);
  int nr = min(r, cur.cr);
  if (nl > nr) return 0;
  if (nl == cur.cl && nr == cur.cr) {
    if (cur.ccount.count(k)) {
      return cur.ccount[k];
    } else {
      return 0;
    }
  }
  int ret = 0;
  for (int i = nl; i <= nr; ++i) {
    if (cur.a[i - cur.cl + cur.l] == k) {
      ++ret;
    }
  }
  return ret;
}
bool update_block(block& cur, int l, int r, int gofront, int nx) {
  int nl = max(l, cur.cl);
  int nr = min(r, cur.cr);
  if (nl > nr) {
    return false;
  }
  if (nl == cur.cl && nr == cur.cr) {
    --cur.ccount[cur.a[cur.l]];
    if (!cur.ccount.count(cur.a[cur.l])) {
      cur.ccount.erase(cur.a[cur.l]);
    }
    ++cur.l;
    ++cur.r;
    if (r == nr) {
      cur.a[cur.r] = gofront;
    } else {
      cur.a[cur.r] = nx;
    }
    ++cur.ccount[cur.a[cur.r]];
    if (cur.r == D * BSIZE) {
      return true;
    } else {
      return false;
    }
  } else {
    if (nr == cur.cr) {
      --cur.ccount[cur.a[nl - cur.cl + cur.l]];
      if (!cur.ccount.count(cur.a[nl - cur.cl + cur.l])) {
        cur.ccount.erase(cur.a[nl - cur.cl + cur.l]);
      }
      for (int i = nl; i < nr; ++i) {
        cur.a[i - cur.cl + cur.l] = cur.a[i - cur.cl + cur.l + 1];
      }
      if (nr == r) {
        cur.a[nr - cur.cl + cur.l] = gofront;
      } else {
        cur.a[nr - cur.cl + cur.l] = nx;
      }
      ++cur.ccount[cur.a[nr - cur.cl + cur.l]];
    } else {
      --cur.ccount[cur.a[nl - cur.cl + cur.l]];
      if (!cur.ccount.count(cur.a[nl - cur.cl + cur.l])) {
        cur.ccount.erase(cur.a[nl - cur.cl + cur.l]);
      }
      for (int i = nl; i < nr; ++i) {
        cur.a[i - cur.cl + cur.l] = cur.a[i - cur.cl + cur.l + 1];
      }
      cur.a[nr - cur.cl + cur.l] = gofront;
      ++cur.ccount[gofront];
    }
  }
  return false;
}
void get_a_from_blocks() {
  int cnt = 0;
  for (int i = 0; i < have.size(); ++i) {
    for (int j = have[i].l; j <= have[i].r; ++j) {
      a[cnt] = have[i].a[j];
      ++cnt;
    }
  }
}
void update_all(int l, int r) {
  int gofront;
  for (int i = 0; i < have.size(); ++i) {
    if (have[i].cl <= l && have[i].cr >= l) {
      gofront = have[i].a[l - have[i].cl + have[i].l];
      break;
    }
  }
  bool need = false;
  for (int i = 0; i < have.size(); ++i) {
    int nx = -1;
    if (i + 1 < have.size()) {
      nx = have[i + 1].a[have[i + 1].l];
    }
    need |= update_block(have[i], l, r, gofront, nx);
  }
  if (need) {
    get_a_from_blocks();
    update_blocks();
  }
}
int get_all(int l, int r, int k) {
  int ret = 0;
  for (int i = 0; i < have.size(); ++i) {
    ret += get_count(have[i], l, r, k);
  }
  return ret;
}
void output() {
  get_a_from_blocks();
  reverse(a, a + n);
  for (int i = 0; i < n; ++i) {
    printf("%d ", a[i]);
  }
  printf("\n");
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  reverse(a, a + n);
  update_blocks();
  int q;
  scanf("%d", &q);
  int lastans = 0;
  for (int i = 0; i < q; ++i) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      --l, --r;
      l = n - 1 - l;
      r = n - 1 - r;
      if (l > r) swap(l, r);
      update_all(l, r);
    } else {
      int l, r, k;
      scanf("%d%d%d", &l, &r, &k);
      l = (l + lastans - 1) % n;
      r = (r + lastans - 1) % n;
      k = (k + lastans - 1) % n + 1;
      l = n - 1 - l;
      r = n - 1 - r;
      if (l > r) swap(l, r);
      lastans = get_all(l, r, k);
      printf("%d\n", lastans);
    }
  }
  return 0;
}
