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: Accepted)

#include <iostream> // cout, endl, cin
#include <iomanip>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#define rep(i,N) for (ll i = 0; i < (N); i++)
#define repone(i,N) for(ll i = 1; i < (N); i++)
#define REP(i,N) for (ll i = (N)-1; i >= 0; i--)
#define FOR(j,i,N) for (ll j = (N)-1; j > (i); j--)
#define repsqrt(i,N) for(ll i = 1; i*i <= (N); i++)
#define loop(i,a,b) for(ll i = a; i < b;i++)
#define ST string
#define vec vector<ll>
#define vecs vector<string>
#define outb(s) cout << fixed << setprecision(15) << s << endl;
#define out(s) cout << s << endl;
#define SZ(x) ((ll)(x).size())
#define Graph vector<vector<ll>>
#define vecb vector<lb>
#define P pair<ll, ll>
#define F first
#define S second
#define vecbool vector<bool>
using ll = long long;
using lb = long double;
using ll = long long;
using lb = long double;
using namespace std;
const ll mod = 1000000007;
const ll zero = 0;
const ll one = 1;
const ll INF = 1e8;
const lb pi = 3.14159265358979;
 
ll nCk(ll N, ll K){
  if(N>1){
    ll kid = 1;
  for(ll i = N; i > N - K; i--){
      kid = kid * i;
    }
    for(ll i = 1; i < K + 1; i++){
        kid = kid / i;
    }
  return kid;
  }
  else{
    return 0;
  }
}
 
ll stair_pow(ll N){// 階乗
  ll sum = 1;
  for(ll i = 1; i <= N; i++){
    sum = sum * i % mod;
  }
  return sum % mod;
}
 
ll gcd(ll p, ll q){
  return q ? gcd(q ,p % q):p;
}
 
ll lcm(ll p, ll q){
  return p / gcd(p, q) * q;
}
 
bool is_prime(ll x){
  if(x <= 1){
    return false;
  }
  for(ll i=2; i * i <= x; i++){
    if(x % i == 0){ 
      return false;
    }
  }
  return true;
}
 
ll sum_of_num(ll num){// 各位の和
  ll dig;
  ll sum = 0;
  while(num){
    dig = num % 10;
    sum = sum + dig;
    num = num / 10;
  }
  return sum;
}
 
ll how_many_break(ll n, ll m){// 何回割れるか
  ll counter = 0;
  while (n % m == 0){
    n = n / m;
    counter++;
  }
  return counter;
}
 
ll many_pow(ll N, ll M){ // NのM乗
  if(M == 0)return 1;
  else{
    ll sum = 1;
    for(ll i = 0; i < M; i++){
      sum *= N;
    }
    return sum;
  }
}
 
ll one_to_i(ll i){ // 1からiまでの和
  if(i < 0){
    return 0;
  }
  else{
    return i*(i+1)/2;
  }
}
 
ll how_many_yaku(ll num){
  ll ans = 0;
   repsqrt(i,num){
    if(num % i == 0){
      ans++;
      if(i != num/i){
        ans++;
      }
    }
  }
  return ans;
}
 
ll digit(ll num){
    ll digit=0;
    while(num!=0){
        num /= 10;
        digit++;
    }
    return digit;
}

ll digitsum(ll num){
  ll dig = 0,sum = 0;
   while(num){
    dig = num % 10;
    sum += dig;
    num /= 10;
  }
  return sum;
}
 
struct UnionFind {
  vec d;
  UnionFind(ll n = 0): d(n,-1) {}
  ll find(ll x) {
    if (d[x] < 0) return x;
    return d[x] = find(d[x]);
  }
  bool unite(ll x, ll y) {
    x = find(x); y = find(y);
    if (x == y) return false;
    if (d[x] > d[y]) swap(x,y);
    d[x] += d[y];
    d[y] = x;
    return true;
  }
  bool same(ll x, ll y) { return find(x) == find(y);}
  int size(ll x) { return -d[find(x)];}
};
 
vecbool seen;
void dfs(const Graph &G, ll v) {
    seen[v] = true;
    for (auto next_v : G[v]) { 
        if (seen[next_v]) continue;
        dfs(G, next_v); 
    }
}

ll LCS(ST s, ST t){
    ll s1 = s.size();
    ll s2 = t.size();
    Graph p(s1 + 1, vec(s2 + 2));
    rep(i,s1 + 1) {
        rep(j,s2 + 1){
            if (i > 0 && j > 0){
                if (s[i-1] == t[j-1]){
                    p[i][j] = max(p[i - 1][j - 1] + 1, max(p[i][j - 1], p[i - 1][j]));
                }
                else {
                    p[i][j] = max(p[i - 1][j], p[i][j - 1]);
                }
            }
            else {
                p[i][j] = 0;
            }
        }
    }
    ll ans = p[s1][s2];
    return ans;
}
 
ll modpow(ll x, ll y){
    ll res = 1;
    rep(i,y){
        res = res * x % mod;
    }
    return res;
}

struct modint{
  ll x;
  modint(ll x = 0) : x((x % mod + mod) % mod){}
  modint operator-() const {
    return modint(-x);
  }
  modint& operator += (const modint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  modint& operator -= (const modint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  modint& operator *= (const modint a) { 
    (x *= a.x) %= mod; 
    return *this;
  }
  modint operator + (const modint a) const { 
    return modint(*this) += a;
  }
  modint operator - (const modint a) const { 
    return modint(*this) -= a;
  }
  modint operator * (const modint a) const { 
    return modint(*this) *= a;
  }
  modint pow(ll t) const {
    if (!t) return 1;
    modint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }
  modint inv() const { 
    return pow(mod - 2);
  }
  modint& operator /= (const modint a) {
    return *this *= a.inv();
  }
  modint operator / (const modint a) const { 
    return modint(*this) /= a;
  }
};
istream& operator >> (istream& is, modint& a) {
  return is >> a.x;
}
ostream& operator << (ostream& os, const modint& a) { 
  return os << a.x;
}

vector<P> prime_factorize(ll N) {
    vector<P> res;
    for (ll a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        ll ex = 0;
        while (N % a == 0) {
            ++ex;
            N /= a;
        }
        res.push_back({a, ex});
    }
    if (N != 1) res.push_back({N, 1});
    return res;
}

/*-----------------------------------------------------------------------------------*/
 
// cout << fixed << setprecision(15)
 
// continue

// count(S.begin(),S.end(),'');
 
// reverse(S.begin(), S.end());
 
// s.substr(8, 8)
 
// sort(p.begin(), p.end());

// reverse(s.begin(), s.end());
 
// sort(p.rbegin(), p.rend()); ← 降順sort.
 
// getline(cin,s);

// transform (s.begin(), s.end(), s.begin(), toupper); ←大文字変換

// transform (s.begin(), s.end(), s.begin(), tolower); ←小文字変換
 
/*-----------------------------------------------------------------------------------*/

int main(){
  ll n,query;
  cin >> n >> query;
  vec a(n + 1), b(n + 1);
  rep(i,n)a[i + 1] = b[i + 1] = n;
  ll black = (n - 2) * (n - 2);
  ll h = n, w = n;
  while(query--) {
    ll t, x;
    cin >> t >> x;
    if (t == 1) {
      if (x < w) {
        loop(i,x,w){
          b[i] = h;
        }
        w = x;
      }
      black -= b[x] - 2;
    }
    else {
      if (x < h) {
        loop(i,x,h) {
          a[i] = w;
        }
        h = x;
      }
      black -= a[x] - 2;
    }
  }
  cout << black << endl;
  return 0;
}