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#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<(int)b;++i)
#define FORD(i,a,b) for(int i=a;i>=(int)b;--i)
#define PB push_back
#define EB emplace_back
#define FI first
#define SE second
#define umap unordered_map
#define uset unordered_set
#define vi vector<int>
#define vvi vector<vi>
#define vll vector<ll>
#define vvll vector<vll>
#define vpii vector<pii>
#define pii pair<int, int>
#define pll pair<ll, ll>
#define ALL(X) (X).begin(),(X).end()
#ifndef DEBUG
#define endl (char)10
#endif
using namespace std;
using ll = long long;
using ld = long double;

template <class T>
istream& operator>> (istream& is, vector<T>& vec){
    FOR(i,0,vec.size()) is >> vec[i];
    return is;
}
template <class T>
ostream& operator<< (ostream& os, vector<T>& vec){
    for(auto& t : vec) os << t << " ";
    return os;
}
template<class T, class U>
ostream& operator<< (ostream& os, const pair<T, U>& p){
    os << p.FI << " " << p.SE;
    return os;
}
template<class T, class U>
istream& operator>> (istream& is, pair<T, U>& p){
    is >> p.FI >> p.SE;
    return is;
}

inline bool is_free(vvi& G, vi& used, int x, int k) {
    return G[x][k] == -1 || used[G[x][k]];
}
void pick(vvi& G, vi& E, vi& used, int x) {
    if (x == -1 || used[x]) return;
    if(is_free(G, used, x, 0) && is_free(G, used, x, 2)){
        E.PB(x);
        used[x] = 1;
        pick(G, E, used, G[x][1]);
        pick(G, E, used, G[x][3]);
    } else if(is_free(G, used, x, 1) && is_free(G, used, x, 3)){
        E.PB(x);
        used[x] = 1;
        pick(G, E, used, G[x][0]);
        pick(G, E, used, G[x][2]);
    }
}

int solve(vvi& G, set<int>& full, vi& used) {
    vi E;
    for(int x : full) pick(G, E, used, x);
    int ans = E.size();
    for(int x : E) used[x] = 0;
    return ans;
}

const int da[4] = {1, 0, -1, 0};
const int db[4] = {0, -1, 0, 1};

int main () {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n, m, K, q;
    cin >> n >> m >> K >> q;
    vector<map<int, int>> V(n + 2);
    // SWNE
    vvi G(K + q, vi(4, -1));
    auto fix_neigh = [&V, &G](int i, int a, int b) {
        FOR(k,0,4) {
            int na = a + da[k];
            int nb = b + db[k];
            auto it = V[na].find(nb);
            if (it != V[na].end()) {
                G[i][k] = it -> SE;
                G[it -> SE][k ^ 2] = i;
            }
        }
    };
    set<int> full;
    vi E(K + q, 0);
    FOR(i,0,K) {
        int a, b;
        cin >> a >> b;
        V[a][b] = i;
        full.insert(i);
        fix_neigh(i, a, b);
    }
    cout << solve(G, full, E) << endl;
    FOR(qid,0,q){
        int a, b;
        cin >> a >> b;
        auto it = V[a].find(b);
        if (it != V[a].end()) {
            full.erase(it -> SE);
            FOR(k,0,4) if(G[it -> SE][k] != -1) G[G[it -> SE][k]][k ^ 2] = -1;
            V[a].erase(b);
        } else {
            V[a][b] = qid + K;
            fix_neigh(qid + K, a, b);
            full.insert(qid + K);
        }
        cout << solve(G, full, E) << endl;
    }
}