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;
const int maxn = 100001;
const int bl = 400;
const int box = maxn / bl + 5;
const int inf = 1 << 29;
const int mod = 1000000007;
const double eps = 1e-8;
int cn[box][maxn], siz[box], v[box][bl * 2];
int n, m, da[maxn], q;
void build() {
  m = 0;
  for (int i = 1; i <= n; i += bl) {
    ++m;
    int r = min(i + bl, n + 1);
    siz[m] = r - i;
    for (int j = i; j < r; j++) {
      v[m][j - i] = da[j];
      cn[m][da[j]]++;
    }
  }
}
void rebuild() {
  int tot = 0;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < siz[i]; j++) {
      da[++tot] = v[i][j];
      cn[i][da[tot]]--;
    }
  }
  build();
}
int del(int p) {
  int s = 0;
  for (int i = 1; i <= m; i++) {
    s += siz[i];
    if (s >= p) {
      int po = siz[i] - 1 - (s - p), res = v[i][po];
      cn[i][res]--;
      for (int j = po; j < siz[i] - 1; j++) v[i][j] = v[i][j + 1];
      siz[i]--;
      return res;
    }
  }
}
void inser(int p, int x) {
  int s = 0;
  for (int i = 1; i <= m; i++) {
    s += siz[i];
    if (s >= p) {
      cn[i][x]++;
      int po = siz[i] - (s - p);
      for (int j = siz[i]; j >= po + 1; j--) v[i][j] = v[i][j - 1];
      v[i][po] = x;
      siz[i]++;
      return;
    }
  }
}
int query(int p, int x) {
  int s = 0, ans = 0;
  for (int i = 1; i <= m; i++) {
    s += siz[i];
    if (s > p) {
      int po = siz[i] - (s - p);
      for (int j = 0; j < po; j++) ans += (v[i][j] == x);
      return ans;
    } else
      ans += cn[i][x];
  }
  return ans;
}
int query(int l, int r, int x) { return query(r, x) - query(l - 1, x); }
void sh(int l, int r) { inser(l - 1, del(r)); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", da + i);
  build();
  scanf("%d", &q);
  int las = 0, ty, l, r, k;
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d", &ty, &l, &r);
    l = (l + las - 1) % n + 1;
    r = (r + las - 1) % n + 1;
    if (l > r) swap(l, r);
    if (ty == 1)
      sh(l, r);
    else {
      scanf("%d", &k);
      k = (k + las - 1) % n + 1;
      las = query(l, r, k);
      printf("%d\n", las);
    }
    if (i % 800 == 0) rebuild();
  }
}