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;
double PI = acos(-1);
double EPS = 1e-7;
int INF = 1000000000;
long long INFLL = 1000000000000000000LL;
int mx[8] = {-1, 1, 0, 0, -1, -1, 1, 1};
int my[8] = {0, 0, -1, 1, -1, 1, -1, 1};
short Count[317][100005];
int Bucket[317][2 * 317];
int Size[317];
int X[100005];
int n, q;
void Build() {
  int cur = 0;
  for (int(a) = (0); (a) <= (n - 1); (a)++) {
    Bucket[cur][Size[cur]] = X[a];
    Count[cur][X[a]]++;
    Size[cur]++;
    if (Size[cur] == 317) cur++;
  }
}
void Rebuild() {
  int c = 0;
  for (int(a) = (0); (a) <= (317 - 1); (a)++) {
    for (int(b) = (0); (b) <= (Size[a] - 1); (b)++) {
      Count[a][Bucket[a][b]]--;
      X[c++] = Bucket[a][b];
    }
    Size[a] = 0;
  }
  Build();
}
void Cetak() {
  int c = 0;
  for (int(a) = (0); (a) <= (317 - 1); (a)++) {
    if (Size[a] == 0) continue;
    cout << "Bucket " << a << " -> ";
    for (int(b) = (0); (b) <= (Size[a] - 1); (b)++) {
      cout << Bucket[a][b] << " ";
    }
    cout << endl;
  }
  cout << endl;
}
int Erase(int p) {
  int cur = 1;
  int d = 0;
  for (int(a) = (0); (a) <= (317 - 1); (a)++) {
    if (Size[a] == 0) continue;
    if (cur <= p && p <= cur + Size[a] - 1) {
      p -= cur;
      d = a;
      break;
    }
    cur += Size[a];
  }
  int ret = Bucket[d][p];
  for (int(a) = (p + 1); (a) <= (Size[d] - 1); (a)++) {
    Bucket[d][a - 1] = Bucket[d][a];
  }
  Count[d][ret]--;
  Size[d]--;
  return ret;
}
void Insert(int p, int k) {
  p--;
  if (p) {
    int cur = 1;
    int d = 0;
    for (int(a) = (0); (a) <= (317 - 1); (a)++) {
      if (Size[a] == 0) continue;
      if (cur <= p && p <= cur + Size[a] - 1) {
        p -= cur;
        d = a;
        break;
      }
      cur += Size[a];
    }
    for (int(a) = (Size[d] - 1); (a) >= (p + 1); (a)--) {
      Bucket[d][a + 1] = Bucket[d][a];
    }
    Bucket[d][p + 1] = k;
    Count[d][k]++;
    Size[d]++;
    if (Size[d] == 2 * 317) Rebuild();
  } else {
    for (int(a) = (Size[0] - 1); (a) >= (0); (a)--) {
      Bucket[0][a + 1] = Bucket[0][a];
    }
    Bucket[0][0] = k;
    Count[0][k]++;
    Size[0]++;
    if (Size[0] == 2 * 317) Rebuild();
  }
}
int Query(int l, int r, int k) {
  int cur = 1;
  int ans = 0;
  for (int(a) = (0); (a) <= (317 - 1); (a)++) {
    if (Size[a] == 0) continue;
    if (l <= cur && cur + Size[a] - 1 <= r) {
      ans += Count[a][k];
    } else {
      int fl = max(l, cur) - cur;
      int fr = min(cur + Size[a] - 1, r) - cur;
      if (fl <= fr) {
        for (int(b) = (fl); (b) <= (fr); (b)++) {
          if (Bucket[a][b] == k) ans++;
        }
      }
    }
    cur += Size[a];
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int(a) = (0); (a) <= (n - 1); (a)++) {
    scanf("%d", &X[a]);
  }
  Build();
  int lastans = 0;
  scanf("%d", &q);
  while (q--) {
    int t, l, r;
    scanf("%d %d %d", &t, &l, &r);
    l = ((l + lastans - 1 + n) % n) + 1;
    r = ((r + lastans - 1 + n) % n) + 1;
    if (l > r) swap(l, r);
    if (t == 1) {
      Insert(l, Erase(r));
    } else {
      int k;
      scanf("%d", &k);
      k = ((k + lastans - 1 + n) % n) + 1;
      lastans = Query(l, r, k);
      printf("%d\n", lastans);
    }
  }
}