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 rt = 374;
int n, A[100010], q, t, l, r, k, lastAns, ans;
int myBlock[100010], blocks = -1, iniBlock[rt], finBlock[rt];
int bu[rt][100010];
deque<int> pseudoList[rt], split1, split2;
void bruteForce() {
  int currentBlock = myBlock[l];
  split1.clear();
  split2.clear();
  for (int i = iniBlock[currentBlock]; i < l; i++) {
    if (pseudoList[currentBlock].empty()) break;
    split1.push_back(pseudoList[currentBlock].front());
    pseudoList[currentBlock].pop_front();
  }
  for (int i = finBlock[currentBlock]; i > r; i--) {
    if (pseudoList[currentBlock].empty()) break;
    split2.push_front(pseudoList[currentBlock].back());
    pseudoList[currentBlock].pop_back();
  }
  int newVal = pseudoList[currentBlock].back();
  pseudoList[currentBlock].push_front(newVal);
  pseudoList[currentBlock].pop_back();
  while (!split1.empty()) {
    pseudoList[currentBlock].push_front(split1.back());
    split1.pop_back();
  }
  while (!split2.empty()) {
    pseudoList[currentBlock].push_back(split2.front());
    split2.pop_front();
  }
  return;
}
void update() {
  if (l == r) return;
  if (myBlock[l] == myBlock[r]) {
    bruteForce();
    return;
  }
  int blockL = myBlock[l], blockR = myBlock[r];
  for (int i = blockL + 1; i <= blockR; i++) {
    bu[i][pseudoList[i - 1].back()]++;
    bu[i - 1][pseudoList[i - 1].back()]--;
    pseudoList[i].push_front(pseudoList[i - 1].back());
    pseudoList[i - 1].pop_back();
  }
  split1.clear();
  split2.clear();
  for (int i = finBlock[blockL]; i > l; i--) {
    if (pseudoList[blockL].empty()) break;
    split1.push_front(pseudoList[blockL].back());
    pseudoList[blockL].pop_back();
  }
  for (int i = finBlock[blockR]; i > r; i--) {
    if (pseudoList[blockR].empty()) break;
    split2.push_front(pseudoList[blockR].back());
    pseudoList[blockR].pop_back();
  }
  int newVal = pseudoList[blockR].back();
  split1.push_front(newVal);
  pseudoList[blockR].pop_back();
  bu[blockL][newVal]++;
  bu[blockR][newVal]--;
  while (!split1.empty()) {
    pseudoList[blockL].push_back(split1.front());
    split1.pop_front();
  }
  while (!split2.empty()) {
    pseudoList[blockR].push_back(split2.front());
    split2.pop_front();
  }
  return;
}
int solve(int L_, int R_) {
  int ans_ = 0;
  if (myBlock[L_] == myBlock[R_]) {
    int b_ = myBlock[L_];
    split1.clear();
    for (int i = finBlock[b_]; i > R_; i--) {
      if (pseudoList[b_].empty()) break;
      split1.push_front(pseudoList[b_].back());
      pseudoList[b_].pop_back();
    }
    for (int i = R_; i >= L_; i--) {
      if (pseudoList[b_].empty()) break;
      split1.push_front(pseudoList[b_].back());
      pseudoList[b_].pop_back();
      if (split1.front() == k) ans_++;
    }
    while (!split1.empty()) {
      pseudoList[b_].push_back(split1.front());
      split1.pop_front();
    }
    return ans_;
  }
  int blockL = myBlock[L_], blockR = myBlock[R_];
  for (int i = blockL + 1; i < blockR; i++) ans_ += bu[i][k];
  return ans_ + solve(L_, finBlock[blockL]) + solve(iniBlock[blockR], R_);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int sig = 0;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    if (i == sig) {
      sig += rt;
      blocks++;
      iniBlock[blocks] = i;
      if (i > 0) finBlock[blocks - 1] = i - 1;
    }
    myBlock[i] = blocks;
    pseudoList[blocks].push_back(A[i]);
    bu[blocks][A[i]]++;
  }
  finBlock[blocks] = n - 1;
  cin >> q;
  while (q--) {
    cin >> t >> l >> r;
    t--;
    l = (l + lastAns - 1) % n;
    r = (r + lastAns - 1) % n;
    if (l > r) swap(l, r);
    if (!t) {
      update();
    } else {
      cin >> k;
      k = ((k + lastAns - 1) % n) + 1;
      ans = solve(l, r);
      cout << ans << "\n";
      lastAns = ans;
    }
  }
  return 0;
}