codeforce ABBYY Cup 3.0

B2. Shave Beaver!

 

The Smart Beaver has recently designed and built an innovative nanotechnologic all-purpose beaver mass shaving machine, "Beavershave 5000". Beavershave 5000 can shave beavers by families! How does it work? Very easily!

There are n beavers, each of them has a unique id from 1 to n. Consider a permutation a1, a2, ..., an of n these beavers. Beavershave 5000 needs one session to shave beavers with ids from x to y (inclusive) if and only if there are such indices i1 < i2 < ... < ik, thatai1 = x, ai2 = x + 1, ..., aik - 1 = y - 1, aik = y. And that is really convenient. For example, it needs one session to shave a permutation of beavers 1, 2, 3, ..., n.

If we can't shave beavers from x to y in one session, then we can split these beavers into groups [x, p1], [p1 + 1, p2], ..., [pm + 1, y](x ≤ p1 < p2 < ... < pm < y), in such a way that the machine can shave beavers in each group in one session. But then Beavershave 5000 needs m + 1 working sessions to shave beavers from x to y.

All beavers are restless and they keep trying to swap. So if we consider the problem more formally, we can consider queries of two types:

• what is the minimum number of sessions that Beavershave 5000 needs to shave beavers with ids from x to y, inclusive?
• two beavers on positions x and y (the beavers ax and ay) swapped.

You can assume that any beaver can be shaved any number of times.

Input

The first line contains integer n — the total number of beavers, 2 ≤ n. The second line contains n space-separated integers — the initial beaver permutation.

The third line contains integer q — the number of queries, 1 ≤ q ≤ 105. The next q lines contain the queries. Each query i looks as pi xiyi, where pi is the query type (1 is to shave beavers from xi to yi, inclusive, 2 is to swap beavers on positions xi and yi). All queries meet the condition: 1 ≤ xi < yi ≤ n.

• to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem B1);
• to get 100 points, you need to solve the problem with constraints: n ≤ 3·105 (subproblems B1+B2).

Note that the number of queries q is limited 1 ≤ q ≤ 105 in both subproblem B1 and subproblem B2.

Output

For each query with pi = 1, print the minimum number of Beavershave 5000 sessions.

Examples

input

5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5

output

2
1
3
5

 题意：

　　给你长度n的序列，m次询问

　　1：x -> y 的花费  满足 每次 选择 以一个a值  能到大其右边任意位置 （即最长连续上升子序列）算一次路径，问从x值到达y值，需要几次

　　2：x,y  交换a[x],a[y];

题解：

　　假设x+1在 x的右边 那么此x的位置值为 1，即任意的区间求和

　　有交换操作，线段树维护a[x],a[y]对序列的影响即可

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 3e5+10, M = 2e5+35000+11, mod = 1e9+7, inf = 0x3fffffff;

int id[N],a[N],n,m,v[N*20];
void update(int i,int ll,int rr,int x,int c) {
        if(ll == rr) {
            v[i] = c;
            return ;
        }
        if(x <= mid) update(ls,ll,mid,x,c);
        else update(rs,mid+1,rr,x,c);
        v[i] = v[ls] + v[rs];
}
int ask(int i,int ll,int rr,int x,int y) {
        if(ll == x && y == rr) {
            return v[i];
        }
        if(y <= mid) return ask(ls,ll,mid,x,y);
        else if(x > mid) return ask(rs,mid+1,rr,x,y);
        else return ask(ls,ll,mid,x,mid) + ask(rs,mid+1,rr,mid+1,y);
}
int main() {
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i) scanf("%d",&a[i]),id[a[i]] = i;
        for(int i = 1; i < n; ++i) {
            if(id[i] > id[i+1]) update(1,1,n,i,1);
        }
        scanf("%d",&m);
        for(int i = 1; i <= m; ++i) {
            int op,x,y;
            scanf("%d%d%d",&op,&x,&y);
            if(op == 1) {
                printf("%d
",ask(1,1,n,x,y-1) + 1);
            } else {
                int tmp1 = a[x];
                int tmp2 = a[y];
                int tt = id[a[x]];
                id[a[x]] = id[a[y]];
                id[a[y]] = tt;
                swap(a[x],a[y]);
                if(tmp1+1 <= n && id[tmp1] > id[tmp1+1]) update(1,1,n,tmp1,1);
                if(tmp1-1 >= 1 && id[tmp1-1] < id[tmp1]) update(1,1,n,tmp1-1,0);

                if(tmp2+1 <= n && id[tmp2] < id[tmp2+1]) update(1,1,n,tmp2,0);
                if(tmp2-1 >= 1 && id[tmp2-1] > id[tmp2]) update(1,1,n,tmp2-1,1);
            }
        }
        return 0;
}