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
#include <bits/stdc++.h>
typedef struct {
  int buf[330];
  int start, end, size, capacity;
} ArrayDeque;
void initArrayDeque(ArrayDeque *a) {
  a->start = a->end = a->size = 0;
  a->capacity = 330;
}
void pushFrontArrayDeque(ArrayDeque *a, int data) {
  if (a->start == 0) a->start = 330;
  a->buf[--a->start] = data;
}
void pushEndArrayDeque(ArrayDeque *a, int data) {
  a->buf[a->end++] = data;
  if (a->end == 330) a->end = 0;
}
int popFrontArrayDeque(ArrayDeque *a) {
  int data = a->buf[a->start++];
  if (a->start == 330) a->start = 0;
  return data;
}
int popEndArrayDeque(ArrayDeque *a) {
  if (a->end == 0) a->end = 330;
  return a->buf[--a->end];
}
ArrayDeque bucket[320];
int count[320][100010], position[100010];
int stack[330], top;
void modify(int l, int r) {
  int i, j, data, rr, ll, pr, pl;
  ArrayDeque *ar, *al;
  pl = position[l];
  pr = position[r];
  al = &bucket[pl];
  ar = &bucket[pr];
  ll = l % 320;
  rr = r % 320;
  top = 0;
  for (i = 0; i < rr; i++) stack[top++] = popFrontArrayDeque(ar);
  data = popFrontArrayDeque(ar);
  count[pr][data]--;
  while (top > 0) pushFrontArrayDeque(ar, stack[--top]);
  for (i = 0; i < ll; i++) stack[top++] = popFrontArrayDeque(al);
  pushFrontArrayDeque(al, data);
  count[pl][data]++;
  while (top > 0) pushFrontArrayDeque(al, stack[--top]);
  for (i = pl; i < pr; i++) {
    data = popEndArrayDeque(&bucket[i]);
    count[i][data]--;
    pushFrontArrayDeque(&bucket[i + 1], data);
    count[i + 1][data]++;
  }
}
int find(int l, int r, int k) {
  int i, j, pl, pr, ll, rr, q1, q2, ans;
  ArrayDeque *al, *ar;
  int *buf, itr;
  pl = position[l];
  pr = position[r];
  al = &bucket[pl];
  ar = &bucket[pr];
  ll = l % 320;
  rr = r % 320;
  ans = q1 = q2 = 0;
  buf = al->buf;
  itr = al->start;
  for (i = 0; i < ll; i++) {
    if (buf[itr++] == k) q1++;
    if (itr == 330) itr = 0;
  }
  buf = ar->buf;
  itr = ar->start;
  for (i = 0; i <= rr; i++) {
    if (buf[itr++] == k) q2++;
    if (itr == 330) itr = 0;
  }
  for (i = pl; i < pr; i++) ans += count[i][k];
  return ans + q2 - q1;
}
void showArrayDeque(ArrayDeque *a) {
  int itr = a->start;
  printf("[");
  while (itr != a->end) {
    printf(" %d", a->buf[itr++]);
    if (itr == 330) itr = 0;
  }
  printf(" ]\n");
}
int main() {
  int i, j, n, m, command, l, r, k, lastAns = 0;
  int bi, data, temp, lvl;
  for (i = 0; i < 320; i++) {
    initArrayDeque(&bucket[i]);
    for (j = 0; j < 100010; j++) count[i][j] = 0;
  }
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &data);
    bi = i / 320;
    position[i] = bi;
    pushEndArrayDeque(&bucket[bi], data);
    count[bi][data]++;
  }
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    scanf("%d %d %d", &command, &l, &r);
    l = (lastAns + l - 1) % n;
    r = (lastAns + r - 1) % n;
    if (l > r) {
      temp = l;
      l = r;
      r = temp;
    }
    if (command == 1)
      modify(l, r);
    else {
      scanf("%d", &k);
      k = ((lastAns + k - 1) % n) + 1;
      lastAns = find(l, r, k);
      printf("%d\n", lastAns);
    }
  }
  return 0;
}
