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;
using ll = long long;
int const N = 100 * 1000 + 16;
int const M = 1000 * 1000 * 1000 + 7;
int n, q;
int sz[320];
int o[320][N];
int block[320][1024];
int a[N];
int blks;
int blk_sz;
void preprocess() {
  blk_sz = sqrt(n) + 1;
  blks = (n + blk_sz - 1) / blk_sz;
}
void build(bool rebuild = true) {
  if (rebuild) {
    int len = 0;
    for (int b = 0; b < blks; ++b) {
      for (int i = 0; i < sz[b]; ++i) {
        a[len] = block[b][i];
        --o[b][a[len]];
        ++len;
      }
      sz[b] = 0;
    }
  }
  for (int i = 0, j = 0, k = 0; i < n; i = j, ++k) {
    while (j - i < blk_sz && j < n) {
      block[k][sz[k]++] = a[j];
      ++o[k][a[j]];
      ++j;
    }
  }
}
int erase_at(int idx_block, int idx) {
  int x = block[idx_block][idx];
  --o[idx_block][x];
  --sz[idx_block];
  for (int i = idx; i < sz[idx_block]; ++i)
    block[idx_block][i] = block[idx_block][i + 1];
  return x;
}
void insert_at(int idx_block, int idx, int x) {
  for (int i = sz[idx_block]; i > idx; --i)
    block[idx_block][i] = block[idx_block][i - 1];
  block[idx_block][idx] = x;
  ++o[idx_block][x];
  ++sz[idx_block];
}
void op1(int l, int r) {
  int first = -1, fdx = -1;
  int second = -1, sdx = -1;
  for (int i = 0, len = 0; i < blks; len += sz[i++]) {
    if (len <= l) first = i, fdx = l - len;
    if (len <= r) second = i, sdx = r - len;
  }
  insert_at(first, fdx, erase_at(second, sdx));
  if (2 * blk_sz < sz[first]) build();
}
int op2(int l, int r, int k) {
  int L = 0;
  int cnt = 0;
  for (int b = 0; b < blks; ++b) {
    int R = L + sz[b] - 1;
    if (l <= L && R <= r)
      cnt += o[b][k];
    else {
      for (int i = 0; i < sz[b]; ++i)
        if (l <= L + i && L + i <= r) cnt += (block[b][i] == k);
    }
    L = R + 1;
  }
  return cnt;
}
void print() {
  for (int b = 0; b < blks; ++b)
    for (int i = 0; i < sz[b]; ++i) cout << block[b][i] << " ";
  cout << "\n";
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  preprocess();
  build(false);
  cin >> q;
  int last = 0;
  while (q--) {
    int t, l, r, k;
    scanf("%d%d%d", &t, &l, &r);
    l = (l + last - 1) % n;
    r = (r + last - 1) % n;
    if (l > r) swap(l, r);
    if (t & 1) {
      op1(l, r);
    } else {
      scanf("%d", &k);
      k = (k + last - 1) % n + 1;
      last = op2(l, r, k);
      printf("%d\n", last);
    }
  }
}