E. Messenger Simulator

Description:
Polycarp is a frequent user of the very popular messenger. He's chatting with his friends all the time. He has n friends, numbered from 1 to n.

Recall that a permutation of size n is an array of size n such that each integer from 1 to n occurs exactly once in this array.

So his recent chat list can be represented with a permutation p of size n. p1 is the most recent friend Polycarp talked to, p2 is the second most recent and so on.

Initially, Polycarp's recent chat list p looks like 1,2,…,n (in other words, it is an identity permutation).

After that he receives m messages, the j-th message comes from the friend aj. And that causes friend aj to move to the first position in a permutation, shifting everyone between the first position and the current position of aj by 1. Note that if the friend aj is in the first position already then nothing happens.

For example, let the recent chat list be p=[4,1,5,3,2]:

if he gets messaged by friend 3, then p becomes [3,4,1,5,2];
if he gets messaged by friend 4, then p doesn't change [4,1,5,3,2];
if he gets messaged by friend 2, then p becomes [2,4,1,5,3].
For each friend consider all position he has been at in the beginning and after receiving each message. Polycarp wants to know what were the minimum and the maximum positions.

Input
The first line contains two integers n and m (1≤n,m≤3e5) — the number of Polycarp's friends and the number of received messages, respectively.

The second line contains m integers a1,a2,…,am (1≤ai≤n) — the descriptions of the received messages.

Output
Print n pairs of integers. For each friend output the minimum and the maximum positions he has been in the beginning and after receiving each message.

Examples
input
5 4
3 5 1 4
output
1 3
2 5
1 4
1 5
1 5
input
4 3
1 2 4
output
1 3
1 2
3 4
1 4
Note
In the first example, Polycarp's recent chat list looks like this:

[1,2,3,4,5]
[3,1,2,4,5]
[5,3,1,2,4]
[1,5,3,2,4]
[4,1,5,3,2]
So, for example, the positions of the friend 2 are 2,3,4,4,5, respectively. Out of these 2 is the minimum one and 5 is the maximum one. Thus, the answer for the friend 2 is a pair (2,5).

In the second example, Polycarp's recent chat list looks like this:

[1,2,3,4]
[1,2,3,4]
[2,1,3,4]
[4,2,1,3]

Problem solving:

1 2 3 4 5

0 0 0 0 1 2 3 4 5

0 0 0 3 1 2 0 4 5
0 0 5 3 1 2 0 4 0
0 1 5 3 0 2 0 4 0
4 1 5 3 0 2 0 0 0



Code:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
#define lowbit(x)    (x & (-x))
int       n, m, mx[maxn], mn[maxn], pos[maxn], mid[maxn];
{
for (; x < maxn; x += lowbit(x))
mid[x] += k;
}
int see(int x)
{
int ans = 0;
for (; x; x -= lowbit(x))
ans += mid[x];
return ans;
}
int main()
{
ios::sync_with_stdio(0);
cin >> n >> m;
for (int i = 1; i <= n; i++)
mx[i] = mn[i] = i, pos[i] = i + m, add(i + m, 1);
for (int i = 0; i < m; i++)
{
int mi; cin >> mi;
mx[mi] = max(mx[mi], see(pos[mi])); mn[mi] = 1;