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 M = 323;
int a[M * M], n;
int cnt[M][100010];
vector<int> val[M];
void build() {
  for (int i = 0; i < M; i++) {
    int st = i * M, ed = min(i * M + M, n);
    if (ed <= st) break;
    val[i].resize(ed - st);
    for (int j = st; j < ed; j++) val[i][j - st] = a[j], cnt[i][a[j]]++;
  }
}
void rebuild() {
  int m = n;
  n = 0;
  for (int i = 0; i < M; i++) {
    int st = i * M, ed = min(i * M + M, m);
    if (st >= ed) break;
    for (int u : val[i]) a[n++] = u, cnt[i][u]--;
    val[i].clear();
  }
  build();
}
void change(int L, int R) {
  int idL = -1, idR = -1;
  for (int i = 0; i < M; i++) {
    if (L <= val[i].size()) {
      idL = i;
      break;
    }
    L -= val[i].size();
  }
  for (int i = 0; i < M; i++) {
    if (R <= val[i].size()) {
      idR = i;
      break;
    }
    R -= val[i].size();
  }
  vector<int> y(val[idR].size() - 1);
  for (int i = 0; i < R - 1; i++) y[i] = val[idR][i];
  int v = val[idR][R - 1];
  for (int i = R; i < val[idR].size(); i++) y[i - 1] = val[idR][i];
  cnt[idR][v]--;
  val[idR] = y;
  vector<int> x(val[idL].size() + 1);
  for (int i = 0; i < L - 1; i++) x[i] = val[idL][i];
  x[L - 1] = v;
  for (int i = L - 1; i < val[idL].size(); i++) x[i + 1] = val[idL][i];
  cnt[idL][v]++;
  val[idL] = x;
}
int calc(int L, int R, int k) {
  int idL = -1, idR = -1;
  for (int i = 0; i < M; i++) {
    if (L <= val[i].size()) {
      idL = i;
      break;
    }
    L -= val[i].size();
  }
  for (int i = 0; i < M; i++) {
    if (R <= val[i].size()) {
      idR = i;
      break;
    }
    R -= val[i].size();
  }
  int ans = 0;
  for (int i = idL + 1; i < idR; i++) ans += cnt[i][k];
  if (idL == idR) {
    for (int i = L; i <= R; i++) ans += val[idL][i - 1] == k;
  } else {
    for (int i = L - 1; i < val[idL].size(); i++) ans += val[idL][i] == k;
    for (int i = 0; i < R; i++) ans += val[idR][i] == k;
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  int q, t, L, R, k, ans = 0, upd = 0;
  build();
  for (cin >> q; q--;) {
    scanf("%d%d%d", &t, &L, &R);
    L = (L + ans - 1) % n + 1;
    R = (R + ans - 1) % n + 1;
    if (L > R) swap(L, R);
    if (t == 1) {
      change(L, R), upd++;
      if (upd == M) rebuild(), upd = 0;
    } else {
      scanf("%d", &k);
      k = (k + ans - 1) % n + 1;
      printf("%d\n", ans = calc(L, R, k));
    }
  }
  return 0;
}