Paul is at the orchestra. The string section is arranged in an r × c rectangular grid and is filled with violinists with the exception of n violists. Paul really likes violas, so he would like to take a picture including at least k of them. Paul can take a picture of any axis-parallel rectangle in the orchestra. Count the number of possible pictures that Paul can take.

Two pictures are considered to be different if the coordinates of corresponding rectangles are different.

Input

The first line of input contains four space-separated integers r, c, n, k (1 ≤ r, c, n ≤ 3000, 1 ≤ k ≤ min(n, 10)) — the number of rows and columns of the string section, the total number of violas, and the minimum number of violas Paul would like in his photograph, respectively.

The next n lines each contain two integers xi and yi (1 ≤ xi ≤ r, 1 ≤ yi ≤ c): the position of the i-th viola. It is guaranteed that no location appears more than once in the input.

Output

Print a single integer — the number of photographs Paul can take which include at least k violas. 

Examples

Input

2 2 1 1
1 2


Output

4


Input

3 2 3 3
1 1
3 1
2 2


Output

1


Input

3 2 3 2
1 1
3 1
2 2


Output

4

Note

We will use '*' to denote violinists and '#' to denote violists.

In the first sample, the orchestra looks as follows: 
    
    
      
    *#  
    **  
    

Paul can take a photograph of just the viola, the 1 × 2 column containing the viola, the 2 × 1 row containing the viola, or the entire string section, for 4 pictures total.

In the second sample, the orchestra looks as follows: 
    
    
      
    #*  
    *#  
    #*  
    

Paul must take a photograph of the entire section.

In the third sample, the orchestra looks the same as in the second sample.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
constexpr long long mod = 1000000007;
const long long INF = mod * mod;
const double eps = 1e-12;
const double pi = acosl(-1.0);
long long mod_pow(long long x, long long n, long long m = mod) {
  if (n < 0) {
    long long res = mod_pow(x, -n, m);
    return mod_pow(res, m - 2, m);
  }
  if (abs(x) >= m) x %= m;
  if (x < 0) x += m;
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % m;
    x = x * x % m;
    n >>= 1;
  }
  return res;
}
struct modint {
  long long n;
  modint() : n(0) { ; }
  modint(long long m) : n(m) {
    if (n >= mod)
      n %= mod;
    else if (n < 0)
      n = (n % mod + mod) % mod;
  }
  operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) {
  a.n += b.n;
  if (a.n >= mod) a.n -= mod;
  return a;
}
modint operator-=(modint& a, modint b) {
  a.n -= b.n;
  if (a.n < 0) a.n += mod;
  return a;
}
modint operator*=(modint& a, modint b) {
  a.n = ((long long)a.n * b.n) % mod;
  return a;
}
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, long long n) {
  if (n == 0) return modint(1);
  modint res = (a * a) ^ (n / 2);
  if (n % 2) res = res * a;
  return res;
}
long long inv(long long a, long long p) {
  return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
modint operator/=(modint& a, modint b) {
  a = a / b;
  return a;
}
const int max_n = 1 << 17;
modint fact[max_n], factinv[max_n];
void init_f() {
  fact[0] = modint(1);
  for (int i = 0; i < max_n - 1; i++) {
    fact[i + 1] = fact[i] * modint(i + 1);
  }
  factinv[max_n - 1] = modint(1) / fact[max_n - 1];
  for (int i = max_n - 2; i >= 0; i--) {
    factinv[i] = factinv[i + 1] * modint(i + 1);
  }
}
modint comb(int a, int b) {
  if (a < 0 || b < 0 || a < b) return 0;
  return fact[a] * factinv[b] * factinv[a - b];
}
modint combP(int a, int b) {
  if (a < 0 || b < 0 || a < b) return 0;
  return fact[a] * factinv[a - b];
}
template <typename T, typename U>
struct SegT {
 private:
  int n;
  vector<T> node;
  vector<U> lazy;
  T et;
  U eu;
  function<T(T, T)> f;
  function<T(T, U, int, int)> g;
  function<U(U, U)> h;

 public:
  SegT(vector<T> ori, T _et, U _eu, function<T(T, T)> _f,
       function<T(T, U, int, int)> _g, function<U(U, U)> _h) {
    int sz = ori.size();
    et = _et, eu = _eu;
    f = _f, g = _g, h = _h;
    n = 1;
    while (n < sz) n <<= 1;
    node.resize(2 * n - 1, et);
    lazy.resize(2 * n - 1, eu);
    for (int i = 0; i < sz; i++) {
      node[i + n - 1] = ori[i];
    }
    for (int i = n - 1 - 1; i >= 0; i--) {
      node[i] = f(node[2 * i + 1], node[2 * i + 2]);
    }
  }
  SegT(int sz, T _et, U _eu, function<T(T, T)> _f,
       function<T(T, U, int, int)> _g, function<U(U, U)> _h) {
    et = _et, eu = _eu;
    f = _f, g = _g, h = _h;
    n = 1;
    while (n < sz) n <<= 1;
    node.resize(2 * n - 1, et);
    lazy.resize(2 * n - 1, eu);
  }
  void eval(int k, int l, int r) {
    node[k] = g(node[k], lazy[k], l, r);
    if (r - l > 1) {
      lazy[2 * k + 1] = h(lazy[k], lazy[2 * k + 1]);
      lazy[2 * k + 2] = h(lazy[k], lazy[2 * k + 2]);
    }
    lazy[k] = eu;
  }
  void add(U x, int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    eval(k, l, r);
    if (r <= a || b <= l) return;
    if (a <= l && r <= b) {
      lazy[k] = h(x, lazy[k]);
      eval(k, l, r);
    } else {
      add(x, a, b, k * 2 + 1, l, (l + r) / 2);
      add(x, a, b, k * 2 + 2, (l + r) / 2, r);
      node[k] = f(node[k * 2 + 1], node[k * 2 + 2]);
    }
  }
  T query(int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    eval(k, l, r);
    if (r <= a || b <= l) return et;
    if (a <= l && r <= b)
      return node[k];
    else {
      T vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
      T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
      return f(vl, vr);
    }
  }
  void update(int loc, T x) {
    int k = 0, l = 0, r = n;
    stack<pair<int, int>> st;
    while (k < n - 1) {
      eval(k, l, r);
      st.push({l, r});
      if (loc < (l + r) / 2) {
        k = 2 * k + 1;
        r = (l + r) / 2;
      } else {
        k = 2 * k + 2;
        l = (l + r) / 2;
      }
    }
    eval(k, l, r);
    st.push({l, r});
    node[k] = x;
    while (k > 0) {
      k = (k - 1) / 2;
      st.pop();
      l = st.top().first, r = st.top().second;
      eval(2 * k + 1, l, (l + r) / 2);
      eval(2 * k + 2, (l + r) / 2, r);
      node[k] = f(node[2 * k + 1], node[2 * k + 2]);
    }
  }
};
int trans[3000][3000];
void solve() {
  int h, w, n, k;
  cin >> h >> w >> n >> k;
  vector<vector<int>> vx(h);
  vector<pair<int, int>> vs;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    vx[a].push_back(b);
    vs.push_back({b, a});
  }
  for (int i = 0; i < h; i++) sort((vx[i]).begin(), (vx[i]).end());
  sort((vs).begin(), (vs).end());
  long long ans = 0;
  for (int i = 0; i < h; i++) {
    vector<pair<int, int>> ps;
    for (int j = 0; j < k; j++) ps.push_back({-1, -1});
    for (int j = 0; j < vs.size(); j++) {
      if (vs[j].second >= i) {
        trans[vs[j].second][vs[j].first] = ps.size();
        ps.push_back(vs[j]);
      }
    }
    for (int j = 0; j < k; j++) ps.push_back({w, -1});
    vector<int> memo(ps.size(), w);
    vector<int> pre(ps.size()), las(ps.size());
    for (int i = 0; i < ps.size(); i++) {
      pre[i] = i - 1;
      las[i] = i + 1;
    }
    vector<bool> exi(ps.size(), true);
    function<int(int)> find_pre = [&](int x) {
      if (exi[pre[x]])
        return pre[x];
      else
        return pre[x] = find_pre(pre[x]);
    };
    function<int(int)> find_las = [&](int x) {
      if (exi[las[x]])
        return las[x];
      else
        return las[x] = find_las(las[x]);
    };
    int sum = 0;
    for (int j = k; j < ps.size() - k; j++) {
      memo[j] = ps[j + k - 1].first;
      sum += memo[j] * (ps[j].first - ps[j - 1].first);
    }
    sum += w * (ps[ps.size() - k].first - ps[ps.size() - k - 1].first - 1);
    int preid[15], lasid[15];
    for (int j = h - 1; j >= i; j--) {
      ans += w * w - sum;
      for (int val : vx[j]) {
        int id = trans[j][val];
        int loc = 0;
        int cur = id;
        for (int t = 0; t < k + 1; t++) {
          lasid[loc] = cur;
          loc++;
          if (t < k) cur = find_las(cur);
        }
        loc = 0, cur = id;
        for (int t = 0; t < k + 1; t++) {
          preid[loc] = cur;
          loc++;
          if (t < k) cur = find_pre(cur);
        }
        for (int t = 0; t < k; t++) {
          sum -= memo[preid[t]] * (ps[preid[t]].first - ps[preid[t + 1]].first);
          memo[preid[t]] = ps[lasid[k - t]].first;
          sum += memo[preid[t]] * (ps[preid[t]].first - ps[preid[t + 1]].first);
        }
        exi[id] = false;
      }
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
