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 NMax = 100005;
const int sqrNMax = 330;
int N;
int Nb[NMax][sqrNMax + 5];
int A[NMax], B[sqrNMax + 5][2 * sqrNMax + 5], Start[sqrNMax + 5], Block[NMax],
    totalB, Size[sqrNMax + 5];
void Read() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
}
void buildB() {
  int currentBlock = 1, cnt = 0;
  Start[1] = 1;
  for (int i = 1; i <= N; i++) {
    ++cnt;
    if (cnt == sqrNMax) {
      ++currentBlock;
      Start[currentBlock] = i;
      cnt = 1;
    }
    ++Size[currentBlock];
    B[currentBlock][cnt] = A[i];
    Nb[A[i]][currentBlock]++;
    Block[i] = currentBlock;
  }
  totalB = currentBlock;
}
int findBlock(int pos) {
  int sum = 0;
  for (int i = 1; i <= totalB; i++) {
    if (sum + Size[i] >= pos) {
      Start[i] = sum + 1;
      return i;
    }
    sum += Size[i];
  }
}
int deleteElem(int block, int pos) {
  int p = pos - Start[block] + 1;
  int res = B[block][p];
  Nb[res][block]--;
  for (int j = p; j < Size[block]; j++) swap(B[block][j], B[block][j + 1]);
  --Size[block];
  return res;
}
void addElem(int block, int pos, int val) {
  int p = pos - Start[block] + 1;
  Nb[val][block]++;
  B[block][++Size[block]] = val;
  for (int j = Size[block] - 1; j >= p; j--) swap(B[block][j], B[block][j + 1]);
}
void makeCircularPerm(int left, int right) {
  int b = findBlock(right);
  int val = deleteElem(b, right);
  b = findBlock(left);
  addElem(b, left, val);
}
void buildA() {
  int cnt = 0;
  for (int i = 1; i <= totalB; i++) {
    for (int j = 1; j <= Size[i]; j++) A[++cnt] = B[i][j];
  }
}
void initialize() {
  for (int i = 1; i <= totalB; i++) {
    for (int j = 1; j <= Size[i]; j++) Nb[B[i][j]][i] = 0;
    Size[i] = 0;
    Start[i] = 0;
  }
}
inline int decode(int nb, int lastans) { return (nb + lastans - 1) % N + 1; }
int Query(int left, int right, int val) {
  int ans = 0, b = findBlock(left), b2 = findBlock(right);
  for (int i = b + 1; i < b2; i++) ans += Nb[val][i];
  int l = left - Start[b] + 1, r = right - Start[b2] + 1;
  if (b == b2) {
    for (int j = l; j <= r; j++)
      if (B[b][j] == val) ++ans;
    return ans;
  }
  for (int j = l; j <= Size[b]; j++)
    if (B[b][j] == val) ans++;
  for (int j = 1; j <= r; j++)
    if (B[b2][j] == val) ++ans;
  return ans;
}
void Solve() {
  int Q;
  scanf("%d", &Q);
  buildB();
  int lastans = 0, cntQ = 0;
  for (int i = 1; i <= Q; i++) {
    int type, l, r, k;
    scanf("%d%d%d", &type, &l, &r);
    l = decode(l, lastans);
    r = decode(r, lastans);
    if (l > r) swap(l, r);
    if (type == 1) {
      ++cntQ;
      makeCircularPerm(l, r);
      if (cntQ == sqrNMax) {
        cntQ = 0;
        buildA();
        initialize();
        buildB();
      }
    } else {
      scanf("%d", &k);
      k = decode(k, lastans);
      lastans = Query(l, r, k);
      printf("%d\n", lastans);
    }
  }
}
int main() {
  Read();
  Solve();
  return 0;
}