#include <bits/stdc++.h>
using namespace std;
using namespace std;
int deg[200100];
set<pair<int, int> > pset;
vector<pair<int, int> > query;
set<int> conn[200100];
bool is_going(int u) { return (pset.find(make_pair(deg[u], u)) != pset.end()); }
int n, m, k;
void delete_adjacent(int u) {
  for (auto v : conn[u]) {
    conn[v].erase(conn[v].find(u));
    if (is_going(v)) {
      pset.erase(pset.find(make_pair(deg[v], v)));
      deg[v]--;
      if (deg[v] < k) {
        delete_adjacent(v);
      } else
        pset.insert(make_pair(deg[v], v));
    }
  }
  conn[u].clear();
}
void print_set() {
  if (!pset.size()) cout << "Set is empty" << endl;
  for (auto p : pset) cout << p.second << "," << p.first << " ";
  cout << endl;
}
vector<int> ans(200100);
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    deg[u]++;
    deg[v]++;
    query.push_back({u, v});
    conn[u].insert(v);
    conn[v].insert(u);
  }
  for (int i = 1; i <= n; i++) {
    pset.insert(make_pair(deg[i], i));
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] < k && is_going(i)) {
      pset.erase(make_pair(deg[i], i));
      delete_adjacent(i);
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    int u = query[i].first;
    int v = query[i].second;
    ans[i + 1] = pset.size();
    if (pset.find(make_pair(deg[u], u)) != pset.end() &&
        pset.find(make_pair(deg[v], v)) != pset.end()) {
      if (deg[u] - 1 >= k && deg[v] - 1 >= k) {
        pset.erase(pset.find(make_pair(deg[u], u)));
        pset.erase(pset.find(make_pair(deg[v], v)));
        conn[u].erase(conn[u].find(v));
        conn[v].erase(conn[v].find(u));
        deg[u]--;
        deg[v]--;
        pset.insert(make_pair(deg[u], u));
        pset.insert(make_pair(deg[v], v));
      } else if (deg[u] <= k) {
        pset.erase(pset.find(make_pair(deg[u], u)));
        deg[u]--;
        delete_adjacent(u);
      } else {
        pset.erase(pset.find(make_pair(deg[v], v)));
        deg[v]--;
        delete_adjacent(v);
      }
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  cout << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.

We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold: 

  * Either this person does not go on the trip, 
  * Or at least k of his friends also go on the trip. 



Note that the friendship is not transitive. That is, if a and b are friends and b and c are friends, it does not necessarily imply that a and c are friends.

For each day, find the maximum number of people that can go on the trip on that day.

Input

The first line contains three integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 2 ⋅ 10^5, 1 ≤ k < n) — the number of people, the number of days and the number of friends each person on the trip should have in the group.

The i-th (1 ≤ i ≤ m) of the next m lines contains two integers x and y (1≤ x, y≤ n, x≠ y), meaning that persons x and y become friends on the morning of day i. It is guaranteed that x and y were not friends before.

Output

Print exactly m lines, where the i-th of them (1≤ i≤ m) contains the maximum number of people that can go on the trip on the evening of the day i.

Examples

Input

4 4 2
2 3
1 2
1 3
1 4


Output

0
0
3
3


Input

5 8 2
2 1
4 2
5 4
5 2
4 3
5 1
4 1
3 2


Output

0
0
0
3
3
4
4
5


Input

5 7 2
1 5
3 2
2 5
3 4
1 2
5 3
1 3


Output

0
0
0
0
3
4
4

Note

In the first example, 

  * 1,2,3 can go on day 3 and 4. 



In the second example, 

  * 2,4,5 can go on day 4 and 5. 
  * 1,2,4,5 can go on day 6 and 7. 
  * 1,2,3,4,5 can go on day 8. 



In the third example, 

  * 1,2,5 can go on day 5. 
  * 1,2,3,5 can go on day 6 and 7. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using namespace std;
int deg[200100];
set<pair<int, int> > pset;
vector<pair<int, int> > query;
set<int> conn[200100];
bool is_going(int u) { return (pset.find(make_pair(deg[u], u)) != pset.end()); }
int n, m, k;
void delete_adjacent(int u) {
  for (auto v : conn[u]) {
    conn[v].erase(conn[v].find(u));
    if (is_going(v)) {
      pset.erase(pset.find(make_pair(deg[v], v)));
      deg[v]--;
      if (deg[v] < k) {
        delete_adjacent(v);
      } else
        pset.insert(make_pair(deg[v], v));
    }
  }
  conn[u].clear();
}
void print_set() {
  if (!pset.size()) cout << "Set is empty" << endl;
  for (auto p : pset) cout << p.second << "," << p.first << " ";
  cout << endl;
}
vector<int> ans(200100);
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    deg[u]++;
    deg[v]++;
    query.push_back({u, v});
    conn[u].insert(v);
    conn[v].insert(u);
  }
  for (int i = 1; i <= n; i++) {
    pset.insert(make_pair(deg[i], i));
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] < k && is_going(i)) {
      pset.erase(make_pair(deg[i], i));
      delete_adjacent(i);
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    int u = query[i].first;
    int v = query[i].second;
    ans[i + 1] = pset.size();
    if (pset.find(make_pair(deg[u], u)) != pset.end() &&
        pset.find(make_pair(deg[v], v)) != pset.end()) {
      if (deg[u] - 1 >= k && deg[v] - 1 >= k) {
        pset.erase(pset.find(make_pair(deg[u], u)));
        pset.erase(pset.find(make_pair(deg[v], v)));
        conn[u].erase(conn[u].find(v));
        conn[v].erase(conn[v].find(u));
        deg[u]--;
        deg[v]--;
        pset.insert(make_pair(deg[u], u));
        pset.insert(make_pair(deg[v], v));
      } else if (deg[u] <= k) {
        pset.erase(pset.find(make_pair(deg[u], u)));
        deg[u]--;
        delete_adjacent(u);
      } else {
        pset.erase(pset.find(make_pair(deg[v], v)));
        deg[v]--;
        delete_adjacent(v);
      }
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  cout << endl;
  return 0;
}

```