#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=(r)-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={1,0,-1,0,1,1,-1,-1,0};
const ll dx[9]={0,-1,0,1,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template <class T>
class SegTree {
    int n;// 葉の数
    vector<T> data;// データを格納するvector
    T def; // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;// 点更新で使う処理

    T find(int a, int b) {
        T val_left = def, val_right = def;
        for (a += (n - 1), b += (n - 1); a < b; a >>= 1, b >>= 1)
        {
            if ((a & 1) == 0){
                val_left = operation(val_left, data[a]);
            }
            if ((b & 1) == 0){
                val_right = operation(data[--b],val_right);
            }
        }
        return operation(val_left, val_right);
    }
    public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
        function<T(T, T)> _update=[](T a,T b){return b;})
        : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }
    void set(int i, T x) { data[i + n - 1] = x; }
    void build() {
        for (int k=n-2;k>=0;k--) data[k] = operation(data[2*k+1],data[2*k+2]);
    }
    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }

    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        //return _query(a, b, 0, 0, n);
        return find(a,b);
    }

    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};


int main(){
    ll h,w,n;cin >> h >> w >> n;
    vvl g(h);rep(i,n){
        ll p;cin >> p;p--;
        ll q;cin >> q;q--;
        g[p].pb(q);
    }
    /*rep(i,h){
        for(auto p:x[i])cout << p <<" ";cout << endl;
    }*/
    SegTree<ll> st(w,0LL,[](ll a,ll b){return a+b;});
    rep(i,w)st.set(i,1);st.build();
    bool flag=true;
    ll ans=0;
    rep(i,h){
        ll left=INF;
        for(auto p:g[i]){
            if(p==0)flag=false;
            st.change(p,0);
            chmin(left,p);
        }
        if(i==0&&left!=INF){
            for(int j=left;j<w;j++)st.change(j,0);
        }
        if(!flag)ans+=st.query(0,w);
        else{
            if(left==INF)ans+=w;
            else ans+=left+st.query(left,w);
        }   
        //cout << ans <<endl;
    }
    cout << ans <<endl;
}

#include <bits/stdc++.h>
using namespace std;
#define LL long long

int main() {
  ios_base::sync_with_stdio(false); cin.tie(NULL);

  int n; LL k; cin >> n >> k;

  LL dp[4][3000005] = {0};

  dp[0][0] = 1;
  
  #if 0
  // A slower way of calculating dp[i][j] is:
  // Will give TLE: https://atcoder.jp/contests/abc200/submissions/22497898

  for (int i = 0; i < 3; i ++) {
    // O(N^2)
    for (int j = 0; j <= 3 * n; j ++) {
      for (int k = 1; k <= n; k ++) {
        dp[i + 1][j + k] += dp[i][j];
      }
    }
  }

  // first of all, we know dp[i][j] > 0 if j <= i * n;
  // next, dp[i][j] will be added to dp[i+1][j+1:j+n+1)
  // so we can basically use dp[i+1][j] as a prefix to dp[i+1][j+1]
  #else

  for (int i = 0; i < 3; i ++) {
    // O(N)
    for (int j = 0; j <= i * n; j ++) {
      dp[i+1][j+1] += dp[i][j];
      dp[i+1][j+n+1] -= dp[i][j]; // remove from accumulation.
    }
    // accumulate / prefix sum
    for (int j = 1; j <= (i + 1) * n; j ++)
      dp[i+1][j] += dp[i+1][j-1];
  }

  #endif
  // we accumulate the counts of sums until we reach k.
  // sum_{i=3}^{x} dp[3][i] <= k. we want to find x.
  int x;
  for (int i = 3; i <= 3 * n; i ++) {
    if (k <= dp[3][i]) { x = i; break; }
    else { k -= dp[3][i]; }
  }

  // finally, we can iterate the beauty of the cake
  for (int i = 1; i <= n; i ++) {
    // i is the beauty.
    // the maximum of the taste we can get is
    // x (the sum) - i (beauty) - n (pop).
    // the minimum is
    // x - i - 1.
    // we should constrain them between [1, n].
    int jmi = max(1, x - i - n), jma = min(n, x - i - 1);

    if (jmi > jma) continue; // in this case i is too small.
    if (k > (jma - jmi + 1)) {
      k -= (jma - jmi + 1); // these are the choices available for taste.
      continue;
    }

    int y = jmi + k - 1;
    int z = x - i - y;
    cout << i << ' ' << y << ' ' << z << '\n';
    return 0;
  }
}