#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n, m, k, deg[N], ans[N];
pair<int, int> edges[N];
set<int> adj[N];
set<pair<int, int> > st;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    edges[i] = {x, y};
    adj[x].insert(y);
    adj[y].insert(x);
  }
  for (int i = 1; i <= n; i++) {
    st.insert({(int)adj[i].size(), i});
  }
  for (int i = m; i >= 1; i--) {
    while (!st.empty() && (*st.begin()).first < k) {
      int node = (*st.begin()).second;
      st.erase(st.begin());
      for (auto ch : adj[node]) {
        st.erase(st.find({(int)adj[ch].size(), ch}));
        adj[ch].erase(node);
        st.insert({(int)adj[ch].size(), ch});
      }
      adj[node].clear();
    }
    ans[i] = (int)st.size();
    if (adj[edges[i].first].find(edges[i].second) !=
        adj[edges[i].first].end()) {
      int v = edges[i].first, u = edges[i].second;
      pair<int, int> pr = {(int)adj[v].size(), v},
                     pr2 = {(int)adj[u].size(), u};
      st.erase(pr);
      st.erase(pr2);
      adj[v].erase(u), adj[u].erase(v);
      pr = {(int)adj[v].size(), v}, pr2 = pr2 = {(int)adj[u].size(), u};
      st.insert(pr);
      st.insert(pr2);
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int N = 3e5 + 5;
int n, m, k, deg[N], ans[N];
pair<int, int> edges[N];
set<int> adj[N];
set<pair<int, int> > st;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    edges[i] = {x, y};
    adj[x].insert(y);
    adj[y].insert(x);
  }
  for (int i = 1; i <= n; i++) {
    st.insert({(int)adj[i].size(), i});
  }
  for (int i = m; i >= 1; i--) {
    while (!st.empty() && (*st.begin()).first < k) {
      int node = (*st.begin()).second;
      st.erase(st.begin());
      for (auto ch : adj[node]) {
        st.erase(st.find({(int)adj[ch].size(), ch}));
        adj[ch].erase(node);
        st.insert({(int)adj[ch].size(), ch});
      }
      adj[node].clear();
    }
    ans[i] = (int)st.size();
    if (adj[edges[i].first].find(edges[i].second) !=
        adj[edges[i].first].end()) {
      int v = edges[i].first, u = edges[i].second;
      pair<int, int> pr = {(int)adj[v].size(), v},
                     pr2 = {(int)adj[u].size(), u};
      st.erase(pr);
      st.erase(pr2);
      adj[v].erase(u), adj[u].erase(v);
      pr = {(int)adj[v].size(), v}, pr2 = pr2 = {(int)adj[u].size(), u};
      st.insert(pr);
      st.insert(pr2);
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
  return 0;
}

```