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 bucketElement {
  int bucketIndex, localPosition, value;
};
int N, A[100000 + 5], freq[317 + 5][100000 + 5];
vector<int> bucket[317 + 5];
void Read() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    int tmp;
    scanf("%d", &tmp);
    bucket[1].push_back(tmp);
  }
}
void Debug() {}
void Rebuild() {
  int j = 1;
  for (int i = 1; i <= 317; i++) {
    for (int q = 0; q < bucket[i].size(); q++) {
      A[j] = bucket[i][q];
      freq[i][A[j]] = 0;
      j++;
    }
    bucket[i].clear();
  }
  int i = 1;
  for (j = 1; j <= N; j++) {
    bucket[i].push_back(A[j]);
    freq[i][A[j]]++;
    if (bucket[i].size() >= 317) {
      i++;
    }
  }
}
bucketElement Find(int pos) {
  for (int i = 1; i <= 317; i++) {
    if (pos > bucket[i].size()) {
      pos -= bucket[i].size();
    } else {
      bucketElement x;
      x.bucketIndex = i;
      x.localPosition = pos - 1;
      x.value = bucket[i][x.localPosition];
      return x;
    }
  }
}
void Cycle(int from, int to) {
  Debug();
  bucketElement bucketA = Find(from);
  bucketElement bucketB = Find(to);
  bucket[bucketB.bucketIndex].erase(bucket[bucketB.bucketIndex].begin() +
                                    bucketB.localPosition);
  freq[bucketB.bucketIndex][bucketB.value]--;
  bucket[bucketA.bucketIndex].insert(
      bucket[bucketA.bucketIndex].begin() + bucketA.localPosition,
      bucketB.value);
  freq[bucketA.bucketIndex][bucketB.value]++;
  Debug();
}
int HowMany(int from, int to, int val) {
  bucketElement bucketA = Find(from);
  bucketElement bucketB = Find(to);
  if (bucketA.bucketIndex == bucketB.bucketIndex) {
    int ans = 0;
    for (int i = bucketA.localPosition; i <= bucketB.localPosition; i++) {
      if (bucket[bucketA.bucketIndex][i] == val) {
        ans++;
      }
    }
    return ans;
  } else {
    int ans = 0;
    for (int i = bucketA.localPosition; i < bucket[bucketA.bucketIndex].size();
         i++) {
      if (bucket[bucketA.bucketIndex][i] == val) {
        ans++;
      }
    }
    for (int i = 0; i <= bucketB.localPosition; i++) {
      if (bucket[bucketB.bucketIndex][i] == val) {
        ans++;
      }
    }
    for (int i = bucketA.bucketIndex + 1; i <= bucketB.bucketIndex - 1; i++) {
      ans += freq[i][val];
    }
    return ans;
  }
}
void Solve() {
  int Q, lastAns = 0, l, lTonos, r, rTonos, k, kTonos, type;
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    if (i % 317 == 0) {
      Rebuild();
    }
    scanf("%d %d %d", &type, &lTonos, &rTonos);
    l = ((lTonos + lastAns - 1) % N) + 1;
    r = ((rTonos + lastAns - 1) % N) + 1;
    if (r < l) {
      swap(l, r);
    }
    if (type == 1) {
      Cycle(l, r);
    } else if (type == 2) {
      scanf("%d", &kTonos);
      k = ((kTonos + lastAns - 1) % N) + 1;
      lastAns = HowMany(l, r, k);
      printf("%d\n", lastAns);
    }
  }
}
int main() {
  Read();
  Solve();
}