CodeForces - 455D

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:

l_i = ((l'_i + lastans - 1) mod n) + 1; r_i = ((r'_i + lastans - 1) mod n) + 1; k_i = ((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 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

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
分块+双端队列
每次更新对每块进行操作：
1.中间的整块：取出最后一个放入下一块的前端；
2.两端的块：左边的块放入给定右边界位置的数字，右端的块删除给定右边界位置的数字
每次询问：
中间的整块直接查询，两边的块遍历可访问的位置；

 #include <cstdio>
 #include <stack>
 #include <cmath>
 #include <queue>
 #include <string>
 #include <queue>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 
 #define lid id<<1
 #define rid id<<1|1
 #define closein cin.tie(0)
 #define scac(a) scanf("%c",&a)
 #define scad(a) scanf("%d",&a)
 #define print(a) printf("%d\n",a)
 #define debug printf("hello world")
 #define form(i,n,m) for(int i=n;i<m;i++)
 #define mfor(i,n,m) for(int i=n;i>m;i--)
 #define nfor(i,n,m) for(int i=n;i>=m;i--)
 #define forn(i,n,m) for(int i=n;i<=m;i++)
 #define scadd(a,b) scanf("%d%d",&a,&b)
 #define memset0(a) memset(a,0,sizeof(a))
 #define scaddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
 #define scadddd(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)
 
 #define INF 0x3f3f3f3f
 #define maxn 100005
 typedef long long ll;
 using namespace std;
 //---------AC(^-^)AC---------\\
 
 deque<int>::iterator it;
 struct node
 {
     deque<int> q;
     int num[maxn];
 }block[];
 int n, m, blo, ans;
 int pos[maxn];
 
 void update(int a, int b)
 {
     if (pos[a] == pos[b])
     {
         it = block[pos[b]].q.begin() + ((b-)%blo);
         int tmp = *it;
         block[pos[b]].q.erase(block[pos[b]].q.begin() + ((b-)%blo));
         block[pos[b]].q.insert(block[pos[a]].q.begin() + ((a-)%blo), tmp);
     }
     else
     {
         it = block[pos[b]].q.begin() + ((b-)%blo);
         int tmp = *it;
         block[pos[b]].num[tmp]--;
         block[pos[b]].q.erase(block[pos[b]].q.begin() + ((b-)%blo));
         for (int i = pos[a]; i < pos[b]; i++) {
             int x = block[i].q.back();
             block[i].q.pop_back();
             block[i].num[x]--;
             block[i + ].q.push_front(x);
             block[i + ].num[x]++;
         }
         block[pos[a]].q.insert(block[pos[a]].q.begin() + ((a-)%blo), tmp);
         block[pos[a]].num[tmp]++;
     }
 }
 void query(int a, int b, int c)
 {
     ans=;
     if (pos[a] == pos[b])
     {
         for (it = block[pos[a]].q.begin() + ((a-)%blo); it <= block[pos[a]].q.begin() + ((b-)%blo); it++)
             if ((*it) == c) ans++;
     }
     else {
         for (int i = pos[a] + ; i < pos[b]; i++)    ans += block[i].num[c];
         for (it = block[pos[a]].q.begin() + ((a-)%blo); it < block[pos[a]].q.end(); it++)
             if ((*it) == c) ans++;
         for (it = block[pos[b]].q.begin(); it <= block[pos[b]].q.begin() + ((b-)%blo); it++)
             if ((*it) == c) ans++;
     }
     printf("%d\n", ans);
 }
 
 int main()
 {
     scad(n);
     blo = sqrt(n);
     if (n%blo) blo++;
     forn(i, , n) pos[i] = (i - )/blo + ;
 
     forn(i, , n) {
         int x;
         scad(x);
         int s = pos[i];
         block[s].q.push_back(x);
         block[s].num[x]++;
     }
     scad(m);
     ans = ;
     while (m--)
     {
         int op;
         scad(op);
         if (op == )
         {
             int l, r;
             scadd(l, r);
             l = (l + ans - ) % n+;
             r = (r + ans - ) % n+;
             if (l > r) swap(l, r);
             update(l, r);
         }
         else
         {
             int l, r, k;
             scaddd(l, r, k);
             l = (l + ans - ) % n + ;
             r = (r + ans - ) % n + ;
             k = (k + ans - ) % n + ;
             if (l > r) swap(l, r);
             query(l, r, k);
         }
     }
     return ;
 }