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 N = 1e5 + 500, M = 320;
int n, q, a[N], bg[N], B, pre[N], suc[N];
int h[N], t[N], L[N], R[N];
short cnt[M][N];
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
inline void del(int x) {
  pre[suc[x]] = pre[x];
  suc[pre[x]] = suc[x];
}
inline void insert(int x, int y) {
  suc[y] = x;
  pre[y] = pre[x];
  suc[pre[x]] = y;
  pre[x] = y;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  if (n <= 4) {
    int last = 0;
    q = read();
    while (q--) {
      int op = read(), l = read(), r = read();
      l = ((l + last - 1) % n) + 1;
      r = ((r + last - 1) % n + 1);
      if (l > r) swap(l, r);
      if (op == 1) {
        int x = a[r];
        for (int i = r; i > l; i--) a[i] = a[i - 1];
        a[l] = x;
      } else {
        int k = read();
        k = ((k + last - 1) % n) + 1;
        last = 0;
        for (int i = l; i <= r; i++)
          if (a[i] == k) last++;
        printf("%d\n", last);
      }
    }
    return 0;
  }
  B = sqrt(n);
  for (int i = 1, id = 0; i <= n; i += B, id++) {
    int lim = min(n, i + B - 1);
    h[id] = lim;
    t[id] = i;
    L[id] = i;
    R[id] = lim;
    for (int j = i; j <= lim; j++) bg[j] = id, cnt[id][a[j]]++;
  }
  for (int i = 1; i <= n; i++) pre[i] = i - 1, suc[i] = i + 1;
  int last = 0;
  q = read();
  while (q--) {
    int op = read(), l = read(), r = read();
    l = ((l + last - 1) % n) + 1;
    r = ((r + last - 1) % n + 1);
    if (l > r) swap(l, r);
    int tl = t[bg[l]], tr = t[bg[r]];
    for (int i = L[bg[l]]; i < l; i++) tl = suc[tl];
    for (int i = L[bg[r]]; i < r; i++) tr = suc[tr];
    if (op == 1) {
      if (l == r) continue;
      if (tr == h[bg[r]]) h[bg[r]] = pre[tr];
      if (tl == t[bg[l]]) t[bg[l]] = tr;
      del(tr);
      insert(tl, tr);
      cnt[bg[r]][a[tr]]--;
      cnt[bg[l]][a[tr]]++;
      for (int i = bg[l]; i < bg[r]; i++) {
        int x = h[i];
        h[i] = pre[h[i]];
        t[i + 1] = x;
        cnt[i][a[x]]--;
        cnt[i + 1][a[x]]++;
      }
    } else {
      int k = read();
      k = ((k + last - 1) % n) + 1;
      if (bg[l] == bg[r]) {
        last = 0;
        for (int i = l; i <= r; i++) last += (a[tl] == k), tl = suc[tl];
        printf("%d\n", last);
        continue;
      }
      last = 0;
      for (int i = bg[l] + 1; i < bg[r]; i++) last += cnt[i][k];
      for (int i = l; i <= R[bg[l]]; i++) last += (a[tl] == k), tl = suc[tl];
      tr = t[bg[r]];
      for (int i = L[bg[r]]; i <= r; i++) last += (a[tr] == k), tr = suc[tr];
      printf("%d\n", last);
    }
  }
}