#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  pair<int, int> vv[m];
  set<int> adj[n];
  int deg[n];
  for (int i = 0; i < n; i++) deg[i] = 0;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    vv[i] = make_pair(x, y);
    deg[x]++;
    deg[y]++;
    adj[x].insert(y);
    adj[y].insert(x);
  }
  int ans[m], good = n;
  bool done[n];
  for (int i = 0; i < n; i++) done[i] = 0;
  set<pair<int, int> > st;
  priority_queue<pair<int, int> > pq;
  for (int i = 0; i < n; i++) pq.push(make_pair(-deg[i], i));
  for (int i = m - 1; i >= 0; i--) {
    while (!pq.empty()) {
      pair<int, int> p = pq.top();
      pq.pop();
      if (-p.first >= k) {
        pq.push(p);
        break;
      }
      if (done[p.second]) continue;
      done[p.second] = true;
      good--;
      int ix = p.second;
      for (set<int>::iterator it = adj[ix].begin(); it != adj[ix].end(); it++) {
        deg[*it]--;
        adj[*it].erase(ix);
        pq.push(make_pair(-deg[*it], *it));
      }
      deg[ix] = 0;
      adj[ix].clear();
    }
    ans[i] = good;
    if (adj[vv[i].first].find(vv[i].second) != adj[vv[i].first].end()) {
      deg[vv[i].first]--;
      pq.push(make_pair(-deg[vv[i].first], vv[i].first));
      deg[vv[i].second]--;
      pq.push(make_pair(-deg[vv[i].second], vv[i].second));
      adj[vv[i].first].erase(vv[i].second);
      adj[vv[i].second].erase(vv[i].first);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  pair<int, int> vv[m];
  set<int> adj[n];
  int deg[n];
  for (int i = 0; i < n; i++) deg[i] = 0;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    vv[i] = make_pair(x, y);
    deg[x]++;
    deg[y]++;
    adj[x].insert(y);
    adj[y].insert(x);
  }
  int ans[m], good = n;
  bool done[n];
  for (int i = 0; i < n; i++) done[i] = 0;
  set<pair<int, int> > st;
  priority_queue<pair<int, int> > pq;
  for (int i = 0; i < n; i++) pq.push(make_pair(-deg[i], i));
  for (int i = m - 1; i >= 0; i--) {
    while (!pq.empty()) {
      pair<int, int> p = pq.top();
      pq.pop();
      if (-p.first >= k) {
        pq.push(p);
        break;
      }
      if (done[p.second]) continue;
      done[p.second] = true;
      good--;
      int ix = p.second;
      for (set<int>::iterator it = adj[ix].begin(); it != adj[ix].end(); it++) {
        deg[*it]--;
        adj[*it].erase(ix);
        pq.push(make_pair(-deg[*it], *it));
      }
      deg[ix] = 0;
      adj[ix].clear();
    }
    ans[i] = good;
    if (adj[vv[i].first].find(vv[i].second) != adj[vv[i].first].end()) {
      deg[vv[i].first]--;
      pq.push(make_pair(-deg[vv[i].first], vv[i].first));
      deg[vv[i].second]--;
      pq.push(make_pair(-deg[vv[i].second], vv[i].second));
      adj[vv[i].first].erase(vv[i].second);
      adj[vv[i].second].erase(vv[i].first);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
  return 0;
}

```