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;
int sqrt_maxn = 505;
const int maxsqrt_n = 205;
const int maxn = 100005;
int a[maxn];
struct node {
  int val;
  node* next;
  node* front;
};
node* front[maxsqrt_n];
node* rail[maxsqrt_n];
int dp[maxsqrt_n][maxn];
node* new_node() {
  static node a[200005];
  static int top = 0;
  return &a[top++];
}
void insert_node(int x, int y) {
  node* t = new_node();
  t->val = y;
  t->next = front[x];
  if (front[x] != 0) front[x]->front = t;
  t->front = 0;
  front[x] = t;
}
void trans_inside(int x, int y) {
  register int j = x / sqrt_maxn;
  register int i;
  register node* t1 = front[j];
  for (i = j * sqrt_maxn; i <= x; i++) {
    t1 = t1->next;
  }
  register node* t2 = t1;
  for (; i <= y; i++) {
    t2 = t2->next;
  }
  t1->front->next = t2;
  t2->front->next = t2->next;
  t2->next->front = t2->front;
  t2->front = t1->front;
  t2->next = t1;
  t1->front = t2;
}
void trans(int x, int y) {
  register int jx = x / sqrt_maxn;
  register int i;
  register node* t1 = front[jx];
  for (i = jx * sqrt_maxn; i <= x; i++) {
    t1 = t1->next;
  }
  int jy = y / sqrt_maxn;
  register node* t2 = front[jy];
  for (i = jy * sqrt_maxn; i <= y; i++) {
    t2 = t2->next;
  }
  t1->front->next = t2;
  t2->front->next = t2->next;
  t2->next->front = t2->front;
  t2->front = t1->front;
  t2->next = t1;
  t1->front = t2;
  dp[jx][t2->val]++;
  dp[jy][t2->val]--;
  for (i = jx; i < jy; i++) {
    t1 = front[i + 1]->next;
    t2 = rail[i]->front;
    t1->front->next = t2;
    t2->front->next = t2->next;
    t2->next->front = t2->front;
    t2->front = t1->front;
    t2->next = t1;
    t1->front = t2;
    dp[i + 1][t2->val]++;
    dp[i][t2->val]--;
  }
}
int calc_inside(int x, int y, int val) {
  int j = x / sqrt_maxn;
  int i;
  node* t1 = front[j];
  for (i = j * sqrt_maxn; i <= x; i++) {
    t1 = t1->next;
  }
  register int ans = 0;
  for (; i <= y; i++) {
    if (t1->val == val) ans++;
    t1 = t1->next;
  }
  if (t1->val == val) ans++;
  return ans;
}
int calc(int x, int y, int val) {
  int j = x / sqrt_maxn;
  int i;
  node* t1 = front[j];
  for (i = j * sqrt_maxn; i <= x; i++) {
    t1 = t1->next;
  }
  int ans = 0;
  for (; t1->val != -2;) {
    if (t1->val == val) ans++;
    t1 = t1->next;
  }
  int jy = y / sqrt_maxn;
  for (i = j + 1; i < jy; i++) {
    ans += dp[i][val];
  }
  t1 = front[jy];
  for (i = jy * sqrt_maxn; i <= y; i++) {
    t1 = t1->next;
    if (t1->val == val) ans++;
  }
  return ans;
}
int main() {
  fprintf(stderr, "%d\n", clock());
  int n;
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  memset(dp, 0, sizeof(dp));
  for (i = 0; i < n; i += sqrt_maxn) {
    register int temp = i / sqrt_maxn;
    front[temp] = 0;
    int j;
    insert_node(temp, -2);
    rail[temp] = front[temp];
    for (j = i + sqrt_maxn - 1; j >= i; j--) {
      if (j >= n) continue;
      insert_node(temp, a[j]);
      dp[temp][a[j]]++;
    }
    insert_node(temp, -1);
  }
  int lastans = 0;
  int q;
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    int oper;
    scanf("%d", &oper);
    if (oper == 1) {
      int x, y;
      scanf("%d%d", &x, &y);
      x = (x - 1 + lastans) % n;
      y = (y - 1 + lastans) % n;
      if (x > y) swap(x, y);
      if (x == y) continue;
      if (x / sqrt_maxn == y / sqrt_maxn) {
        trans_inside(x, y);
      } else {
        trans(x, y);
      }
    } else {
      int x, y, z;
      scanf("%d%d%d", &x, &y, &z);
      x = (x - 1 + lastans) % n;
      y = (y - 1 + lastans) % n;
      z = (z - 1 + lastans) % n;
      if (x > y) swap(x, y);
      if (x / sqrt_maxn == y / sqrt_maxn) {
        lastans = calc_inside(x, y, z);
      } else {
        lastans = calc(x, y, z);
      }
      printf("%d\n", lastans);
    }
  }
  fprintf(stderr, "%d\n", clock());
  return 0;
}