Problem:
Score : 600 points

Problem Statement

There is a grid with N rows and N columns of squares. Let (i, j) be the square at the i-th row from the top and the j-th column from the left.

Each of the central (N-2) \times (N-2) squares in the grid has a black stone on it.
Each of the 2N - 1 squares on the bottom side and the right side has a white stone on it.

Q queries are given. We ask you to process them in order.
There are two kinds of queries. Their input format and description are as follows:

1 x: Place a white stone on (1, x). After that, for each black stone between (1, x) and the first white stone you hit if you go down from (1, x), replace it with a white stone.

2 x: Place a white stone on (x, 1). After that, for each black stone between (x, 1) and the first white stone you hit if you go right from (x, 1), replace it with a white stone.

How many black stones are there on the grid after processing all Q queries?

Constraints

3 \leq N \leq 2\times 10^5

0 \leq Q \leq \min(2N-4,2\times 10^5)

2 \leq x \leq N-1

Queries are pairwise distinct.

Input

Input is given from Standard Input in the following format:

N Q
Query_1
\vdots
Query_Q

Output

Print how many black stones there are on the grid after processing all Q queries.

Sample Input 1

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

Sample Output 1

1

After each query, the grid changes in the following way:

Sample Input 2

200000 0

Sample Output 2

39999200004

Sample Input 3

176527 15
1 81279
2 22308
2 133061
1 80744
2 44603
1 170938
2 139754
2 15220
1 172794
1 159290
2 156968
1 56426
2 77429
1 97459
2 71282

Sample Output 3

31159505795

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <algorithm>
#include <utility>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <functional>

using namespace std;

#define fst first
#define scd second
#define PB push_back
#define MP make_pair
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define omajinai ios::sync_with_stdio(false);cin.tie(0)
#define rep(i,x) for(int i=0;i<(int)(x);++i)
#define rep1(i,x) for(int i=1;i<=(int)(x);++i)
#define rrep(i,x) for(int i=(int)x-1;i>=0;--i)
#define rrep1(i,x) for(int i=(int)x;i>=1;--i)

using ll=long long;
using ld=long double;
using vi=vector<int>;
using vvi=vector<vi>;
using pii=pair<int, int>;
using vpii=vector<pii>;

template<class T,class U>ostream&operator<<(ostream&os,const pair<T,U>p){os<<"("<<p.fst<<", "<<p.scd<<")";return os;}
template<class T>ostream&operator<<(ostream&os,const vector<T>v){rep(i,v.size()){if(i)os<<", ";os<<v[i];}return os;}
template<typename T>T&max(T&a,T&b){if(a>=b)return a;return b;}
template<typename T>T&min(T&a,T&b){if(a < b)return a;return b;}
template<typename T>bool chmax(T&a,T b){if(a < b){a=b;return true;}return false;}
template<typename T>bool chmin(T&a,T b){if(a > b){a=b;return true;}return false;}
constexpr ll TEN(ll n){return n==0?1:10ll*TEN(n-1);}

constexpr ll inf = TEN(9)+5;
constexpr ll linf = 3*TEN(18);

ll N, Q;
int vx[200010], vy[200010];
int xid[200010], yid[200010];
int latest_x[200010], latest_y[200010];

signed main()
{
    cin >> N >> Q;

    ll ans = (N - 2) * (N - 2);
    set<int> X, Y;
    X.insert(N), Y.insert(N);
    bool first_x = true, first_y = true;
    rep(i, Q) {
        int t; scanf("%d", &t);

        if (t == 1) { // x
            int x; scanf("%d", &x);
            xid[x] = i;
            int nxt_x = *X.lower_bound(x);
            ans -= latest_y[xid[nxt_x]];
            latest_x[yid[latest_y[xid[nxt_x]]]] = x;
            X.insert(x);
        } else { // y
            int y; scanf("%d", &y);
            yid[y] = i;
            int nxt_y = *Y.lower_bound(y);
            ans -= latest_x[yid[nxt_y]];
            latest_y[xid[latest_x[yid[nxt_y]]]] = y;
            Y.insert(y);
        }
    }

    if (ans < 0 ) puts("1");
    else cout << ans << endl;
}