# Serega and Fun

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) modn) + 1; ri = ((r i + lastans - 1) modn) + 1; ki = ((k i + lastans - 1) modn) + 1.

Where lastans is the last reply to the query of the 2-nd type (initially, lastans = 0). If after transformation l i is greater than r i, you must swap these values.

Output

For each query of the 2-nd type print the answer on a single line.

Examples

input

76 6 2 7 4 2 571 3 62 2 4 22 2 4 72 2 2 51 2 61 1 42 1 7 3

output

2100

input

88 4 2 2 7 7 8 881 8 82 8 1 71 8 11 7 32 8 8 31 1 41 2 71 4 5

output

20

Solution:

#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>

using namespace std;

const int N = 200010;
const int MAGIC = 474;

int len;
int a[N];

int cnt, from[N], to[N];
vector <int> where[N];

void init() {
cnt = 1;
from[0] = 1;
to[0] = len;
for (int i = 1; i <= len; i++) {
where[i].clear();
}
for (int i = 1; i <= len; i++) {
where[a[i]].push_back(i);
}
}

int get_count(int k, int from, int to) {
int sz = where[k].size();
if (sz == 0) {
return 0;
}
int low = 0, high = sz;
while (low < high) {
int mid = (low + high) >> 1;
if (where[k][mid] < from) {
low = mid + 1;
} else {
high = mid;
}
}
int X = low;
low = -1, high = sz - 1;
while (low < high) {
int mid = (low + high + 1) >> 1;
if (where[k][mid] > to) {
high = mid - 1;
} else {
low = mid;
}
}
int Y = low;
return (Y - X + 1);
}

int new_from[N], new_to[N];

int bounds[4], bound_id;

void cut(int b0, int b1, int b2) {
bounds[0] = b0;
bounds[1] = b1;
bounds[2] = b2;
bounds[3] = 123456789;
bound_id = 0;
int new_cnt = 0;
int sum = 0;
for (int i = 0; i < cnt; i++) {
int new_sum = sum + (to[i] - from[i] + 1);
if (new_sum <= bounds[bound_id]) {
if (new_sum == bounds[bound_id]) {
bound_id++;
}
new_from[new_cnt] = from[i];
new_to[new_cnt] = to[i];
new_cnt++;
sum = new_sum;
continue;
}
int get = bounds[bound_id] - sum;
new_from[new_cnt] = from[i];
new_to[new_cnt] = from[i] + get - 1;
new_cnt++;
from[i] += get;
sum += get;
bound_id++;
i--;
}
cnt = new_cnt;
for (int i = 0; i < cnt; i++) {
from[i] = new_from[i];
to[i] = new_to[i];
}
}

int new_a[N];

int main() {
scanf("%d", &len);
for (int i = 1; i <= len; i++) {
scanf("%d", a + i);
}
init();
int tt;
scanf("%d", &tt);
int lastans = 0;
for (int qq = 1; qq <= tt; qq++) {
int com;
scanf("%d", &com);
if (com == 2) {
int ll, rr, kk;
scanf("%d %d %d", &ll, &rr, &kk);
ll = ((ll + lastans - 1) % len) + 1;
rr = ((rr + lastans - 1) % len) + 1;
kk = ((kk + lastans - 1) % len) + 1;
if (ll > rr) {
swap(ll, rr);
}
int ans = 0;
int sum = 0;
for (int i = 0; i < cnt; i++) {
int x = from[i], y = to[i];
int begin = sum + 1, end = sum + (y - x + 1);
if (ll > begin) {
x += ll - begin;
}
if (rr < end) {
y -= end - rr;
}
if (x <= y) {
ans += get_count(kk, x, y);
}
sum = end;
}
lastans = ans;
printf("%d\n", ans);
} else {
int ll, rr;
scanf("%d %d", &ll, &rr);
ll = ((ll + lastans - 1) % len) + 1;
rr = ((rr + lastans - 1) % len) + 1;
if (ll > rr) {
swap(ll, rr);
}
if (ll == rr) {
continue;
}
cut(ll - 1, rr - 1, rr);
int first = -1;
int sum = 0;
for (int i = 0; i < cnt; i++) {
if (sum == ll - 1) {
first = i;
}
sum += to[i] - from[i] + 1;
if (sum == rr) {
int p = from[i];
for (int j = i; j > first; j--) {
from[j] = from[j - 1];
to[j] = to[j - 1];
}
from[first] = to[first] = p;
break;
}
}
}
if (cnt > MAGIC) {
int pos = 0;
for (int i = 0; i < cnt; i++) {
for (int j = from[i]; j <= to[i]; j++) {
new_a[++pos] = a[j];
}
}
for (int i = 1; i <= len; i++) {
a[i] = new_a[i];
}
init();
}
}
return 0;
}