#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, m, k;
  cin >> n >> m >> k;
  vector<pair<long long int, long long int> > edges(m);
  set<long long int> adj[n];
  for (int i = 0; i < m; ++i) {
    long long int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].insert(v);
    adj[v].insert(u);
    edges[i].first = u;
    edges[i].second = v;
  }
  set<pair<long long int, long long int> > myset;
  set<pair<long long int, long long int> >::iterator it;
  for (int i = 0; i < n; ++i) {
    myset.insert({adj[i].size(), i});
  }
  set<long long int>::iterator itt;
  vector<long long int> ans(m, 0);
  for (int i = m - 1; i > -1; --i) {
    while (!myset.empty() && myset.begin()->first < k) {
      long long int ind = myset.begin()->second;
      myset.erase(myset.begin());
      for (auto v : adj[ind]) {
        it = myset.find({adj[v].size(), v});
        itt = adj[v].find(ind);
        if (it != myset.end() && itt != adj[v].end()) {
          myset.erase(it);
          adj[v].erase(itt);
          myset.insert({adj[v].size(), v});
        }
      }
    }
    ans[i] = myset.size();
    long long int u = edges[i].first, v = edges[i].second;
    it = myset.find({adj[u].size(), u});
    itt = adj[u].find(v);
    if (it != myset.end() && itt != adj[u].end()) {
      myset.erase(it);
      adj[u].erase(itt);
      myset.insert({adj[u].size(), u});
    }
    it = myset.find({adj[v].size(), v});
    itt = adj[v].find(u);
    if (it != myset.end() && itt != adj[v].end()) {
      myset.erase(it);
      adj[v].erase(itt);
      myset.insert({adj[v].size(), v});
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << ans[i] << "\n";
  }
  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);
  cout.tie(0);
  ;
  long long int n, m, k;
  cin >> n >> m >> k;
  vector<pair<long long int, long long int> > edges(m);
  set<long long int> adj[n];
  for (int i = 0; i < m; ++i) {
    long long int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].insert(v);
    adj[v].insert(u);
    edges[i].first = u;
    edges[i].second = v;
  }
  set<pair<long long int, long long int> > myset;
  set<pair<long long int, long long int> >::iterator it;
  for (int i = 0; i < n; ++i) {
    myset.insert({adj[i].size(), i});
  }
  set<long long int>::iterator itt;
  vector<long long int> ans(m, 0);
  for (int i = m - 1; i > -1; --i) {
    while (!myset.empty() && myset.begin()->first < k) {
      long long int ind = myset.begin()->second;
      myset.erase(myset.begin());
      for (auto v : adj[ind]) {
        it = myset.find({adj[v].size(), v});
        itt = adj[v].find(ind);
        if (it != myset.end() && itt != adj[v].end()) {
          myset.erase(it);
          adj[v].erase(itt);
          myset.insert({adj[v].size(), v});
        }
      }
    }
    ans[i] = myset.size();
    long long int u = edges[i].first, v = edges[i].second;
    it = myset.find({adj[u].size(), u});
    itt = adj[u].find(v);
    if (it != myset.end() && itt != adj[u].end()) {
      myset.erase(it);
      adj[u].erase(itt);
      myset.insert({adj[u].size(), u});
    }
    it = myset.find({adj[v].size(), v});
    itt = adj[v].find(u);
    if (it != myset.end() && itt != adj[v].end()) {
      myset.erase(it);
      adj[v].erase(itt);
      myset.insert({adj[v].size(), v});
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << ans[i] << "\n";
  }
  return 0;
}

```