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 st {
  deque<int> q;
  int cnt[100004];
  int use[100004];
  int u_s = 0;
  st() {
    memset(cnt, 0, sizeof(cnt));
    memset(use, 0, sizeof(use));
    q.clear();
  }
  void push_back(int x) {
    q.push_back(x);
    if (use[x] != u_s) {
      use[x] = u_s;
      cnt[x] = 0;
    }
    cnt[x]++;
  }
  void erase(int i) {
    cnt[q[i]]--;
    q.erase(i + q.begin(), i + 1 + q.begin());
  }
  void insert(int i, int x) {
    if (use[x] != u_s) {
      use[x] = u_s;
      cnt[x] = 0;
    }
    cnt[x]++;
    q.insert(q.begin() + i, x);
  }
  void clear() {
    q.clear();
    u_s++;
  }
  int size() { return q.size(); }
};
st b[320];
int a[100004];
int unit;
int n;
int fi;
int bit[320];
int las = 0;
void add(int i, int x) {
  i++;
  while (i < 320) {
    bit[i] += x;
    i += i & -i;
  }
}
int sum(int i) {
  i++;
  int r = 0;
  while (i) {
    r += bit[i];
    i -= i & -i;
  }
  return r;
}
pair<int, int> belong(int i) {
  int mint = 0;
  int maxt = n / unit;
  maxt++;
  while (mint + 1 < maxt) {
    int mid = (mint + maxt) >> 1;
    if (sum(mid) >= i) {
      maxt = mid;
    } else {
      mint = mid;
    }
  }
  pair<int, int> r;
  if (mint != maxt) {
    if (sum(mint) >= i) {
      r.first = mint;
    } else {
      r.first = maxt;
    }
  }
  if (r.first == 0) {
    r.second = i - 1;
    return r;
  } else {
    r.second = i - sum(r.first - 1);
    r.second--;
    return r;
  }
  return r;
}
void re() {
  memset(bit, 0, sizeof(bit));
  int s = 0;
  for (int i = 0; i <= (n) / unit; i++) {
    for (int j = 0; j < b[i].size(); j++) {
      a[s] = b[i].q[j];
      s++;
    }
    b[i].clear();
  }
  for (int i = 0; i < n; i++) {
    b[i / unit].push_back(a[i]);
    add(i / unit, 1);
  }
}
int main() {
  scanf("%d", &n);
  unit = sqrt(100004);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i / unit].push_back(a[i]);
    add(i / unit, 1);
  }
  int q;
  scanf("%d", &q);
  int q2 = 0;
  int uni = sqrt(100004);
  if (uni == 0) {
    uni++;
  }
  uni = max(uni, unit);
  int lastans = 0;
  for (int i = 0; i < q; i++) {
    if (i && (i % uni == 0)) {
      re();
    }
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      l += lastans - 1;
      l %= n;
      l++;
      r += lastans - 1;
      r %= n;
      r++;
      if (l > r) {
        swap(l, r);
      }
      if (l == r) {
        continue;
      }
      pair<int, int> bb = belong(l);
      pair<int, int> c = belong(r);
      int val = b[c.first].q[c.second];
      b[c.first].erase(c.second);
      b[bb.first].insert(bb.second, val);
      add(c.first, -1);
      add(bb.first, 1);
    } else {
      int l, r, k;
      scanf("%d%d%d", &l, &r, &k);
      l += lastans - 1;
      l %= n;
      l++;
      r += lastans - 1;
      r %= n;
      r++;
      k += lastans - 1;
      k %= n;
      k++;
      if (l > r) {
        swap(l, r);
      }
      pair<int, int> c = belong(l);
      pair<int, int> d = belong(r);
      if (c.first == d.first) {
        int ans = 0;
        for (int j = c.second; j <= d.second; j++) {
          if (b[c.first].q[j] == k) {
            ans++;
          }
        }
        lastans = ans;
        printf("%d\n", ans);
        continue;
      }
      int ans = 0;
      for (int j = c.second; j < b[c.first].q.size(); j++) {
        if (b[c.first].q[j] == k) {
          ans++;
        }
      }
      for (int j = 0; j <= d.second; j++) {
        if (b[d.first].q[j] == k) {
          ans++;
        }
      }
      for (int i = c.first + 1; i < d.first; i++) {
        if (b[i].use[k] == b[i].u_s) {
          ans += b[i].cnt[k];
        }
      }
      printf("%d\n", ans);
      lastans = ans;
    }
  }
  return 0;
}