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 ll = long long;
ll MAX_EL = 1e5 + 10;
ll c = 600, C = 1000;
ll count = MAX_EL / c + 10;
ll n, q, cntBlocks;
std::vector<std::vector<ll>> m(count, std::vector<ll>(MAX_EL, 0));
std::vector<std::vector<ll>> vals(count, std::vector<ll>(c + C + 10, 0));
std::vector<std::vector<ll>> wasVals(count, std::vector<ll>(c + C + 10, 0));
std::vector<ll> szVals(count, 0);
std::vector<ll> szWasVals(count, 0);
std::vector<ll> a(MAX_EL, 0);
std::vector<ll> b(MAX_EL, 0);
void Build() {
  for (ll i = 1; i <= n; i += c) {
    ll r = std::min(n, i + c - 1);
    for (ll j = i; j <= r; ++j) {
      ++m[cntBlocks][a[j]];
      vals[cntBlocks][szVals[cntBlocks]++] = a[j];
      wasVals[cntBlocks][szWasVals[cntBlocks]++] = a[j];
    }
    cntBlocks++;
  }
}
void Rebuild() {
  ll buff = 0;
  for (ll i = 0; i < cntBlocks; ++i) {
    for (ll j = 0; j < szVals[i]; ++j) b[++buff] = vals[i][j];
    szVals[i] = 0;
    for (ll j = 0; j < szWasVals[i]; ++j) {
      m[i][wasVals[i][j]] = 0;
    }
    szWasVals[i] = 0;
  }
  cntBlocks = 0;
  for (ll i = 1; i <= n; ++i) a[i] = b[i];
  Build();
}
ll Erase(ll id) {
  ll sum = 0;
  for (ll i = 0; i < cntBlocks; ++i) {
    sum += szVals[i];
    if (sum >= id) {
      ll pos = szVals[i] - (sum - id) - 1;
      m[i][vals[i][pos]]--;
      ll res = vals[i][pos];
      for (ll it = pos; it < szVals[i]; ++it) vals[i][it] = vals[i][it + 1];
      --szVals[i];
      return res;
    }
  }
  return 0;
}
void Insert(ll id, ll k) {
  ll sum = 0;
  for (ll i = 0; i < cntBlocks; ++i) {
    sum += szVals[i];
    if (sum >= id) {
      wasVals[i][szWasVals[i]++] = k;
      ++m[i][k];
      ll pos = id - (sum - szVals[i]);
      for (ll it = szVals[i]; it > pos; it--) vals[i][it] = vals[i][it - 1];
      vals[i][pos] = k;
      ++szVals[i];
      break;
    }
  }
}
void Change(ll l, ll r) {
  if (l == r) {
    return;
  }
  ll x = Erase(r);
  Insert(l - 1, x);
}
ll T_Counter(ll r, ll k) {
  ll sum = 0, res = 0;
  for (ll i = 0; i < cntBlocks; ++i) {
    sum += szVals[i];
    if (sum <= r) {
      res += m[i][k];
    } else {
      sum -= szVals[i];
      ll it = 0;
      while (sum < r) {
        if (vals[i][it] == k) ++res;
        ++sum;
        ++it;
      }
      break;
    }
  }
  return res;
}
ll Counter(ll l, ll r, ll k) { return T_Counter(r, k) - T_Counter(l - 1, k); }
int main() {
  std::cin >> n;
  for (int i = 1; i <= n; ++i) std::cin >> a[i];
  Build();
  std::cin >> q;
  ll z = 1;
  ll lastans = 0;
  for (ll _ = 1; _ <= q; ++_) {
    ll type, l, r, k;
    std::cin >> type;
    if (type == 1) {
      std::cin >> l >> r;
      l = ((l + lastans - 1 + n) % n) + 1;
      r = ((r + lastans - 1 + n) % n) + 1;
      if (l > r) std::swap(l, r);
      Change(l, r);
      ++z;
    } else {
      std::cin >> l >> r >> k;
      l = ((l + lastans - 1 + n) % n) + 1;
      r = ((r + lastans - 1 + n) % n) + 1;
      k = ((k + lastans - 1 + n) % n) + 1;
      if (l > r) std::swap(l, r);
      lastans = Counter(l, r, k);
      std::cout << lastans << '\n';
    }
    if (z % C == 0) Rebuild(), z++;
  }
  return 0;
}