A. Berstagram

Polycarp recently signed up to a new social network Berstagram. He immediately published n posts there. He assigned numbers from 1 to n to all posts and published them one by one. So, just after publishing Polycarp's news feed contained posts from 1 to n — the highest post had number 1, the next one had number 2, ..., the lowest post had number n.

After that he wrote down all likes from his friends. Likes were coming consecutively from the 1-st one till the m-th one. You are given a sequence a1,a2,…,am (1≤aj≤n), where aj is the post that received the j-th like.

News feed in Berstagram works in the following manner. Let's assume the j-th like was given to post aj. If this post is not the highest (first) one then it changes its position with the one above. If aj is the highest post nothing changes.

For example, if n=3, m=5 and a=[3,2,1,3,3], then Polycarp's news feed had the following states:

before the first like: [1,2,3];
after the first like: [1,3,2];
after the second like: [1,2,3];
after the third like: [1,2,3];
after the fourth like: [1,3,2];
after the fifth like: [3,1,2].
Polycarp wants to know the highest (minimum) and the lowest (maximum) positions for each post. Polycarp considers all moments of time, including the moment "before all likes".

Input
The first line contains two integer numbers n and m (1≤n≤105, 1≤m≤4⋅105) — number of posts and number of likes.

The second line contains integers a1,a2,…,am (1≤aj≤n), where aj is the post that received the j-th like.

Output
Print n pairs of integer numbers. The i-th line should contain the highest (minimum) and the lowest (maximum) positions of the i-th post. You should take into account positions at all moments of time: before all likes, after each like and after all likes. Positions are numbered from 1 (highest) to n (lowest).

Examples input Copy

3 5
3 2 1 3 3

output Copy

1 2
2 3
1 3

input Copy

10 6
7 3 5 7 3 6

output Copy

1 2
2 3
1 3
4 7
4 5
6 7
5 7
8 8
9 9
10 10
思路：定义两个数组，分别存放不断改变的当前数和某数的位置。然后模拟。QAQ 

#include<bits/stdc++.h>

using namespace std;
#define int long long
#define N 1000500
struct str{
    int L,R;
}st[N];
int arr[N];
int a[N];// 当前的数 
int b[N];// 某数的位置 
signed main(){
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        st[i].L=i;
        st[i].R=i;
        a[i]=i;
        b[i]=i;
    }
    for(int i=1;i<=m;i++)
        scanf("%lld",&arr[i]);
    for(int i=1;i<=m;i++){
        if(b[arr[i]]!=1){
            int weizhi=b[arr[i]];
            int temp=a[weizhi-1];//前面一个数
            swap(a[weizhi],a[weizhi-1]); 
            b[arr[i]]=b[arr[i]]-1;
            b[temp]=b[temp]+1;
            st[arr[i]].L=min(st[arr[i]].L,b[arr[i]]);
            st[arr[i]].R=max(st[arr[i]].R,b[arr[i]]);
            st[temp].L=min(st[temp].L,b[temp]);
            st[temp].R=max(st[temp].R,b[temp]);
        }
    }
    for(int i=1;i<=n;i++){
        cout<<st[i].L<<" "<<st[i].R<<'\n';
    }
    return 0;
}