#include <bits/stdc++.h>
using namespace std;
const int module = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<set<int>> to(n, set<int>());
  vector<pair<int, int>> edges(m);
  vector<int> nbr(n);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    nbr[a]++;
    nbr[b]++;
    to[a].insert(b);
    to[b].insert(a);
    edges[i] = {a, b};
  }
  vector<bool> vis(n);
  int all = n;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      if (nbr[i] < k) {
        queue<int> q;
        q.push(i);
        vis[i] = true;
        all--;
        while (!q.empty()) {
          int cur = q.front();
          q.pop();
          for (int t : to[cur]) {
            nbr[t]--;
            to[t].erase(cur);
            if (nbr[t] < k && !vis[t]) {
              q.push(t);
              vis[t] = true;
              all--;
            }
          }
        }
      }
    }
  }
  queue<int> q;
  vector<int> answer;
  answer.push_back(all);
  for (int i = m - 1; i > 0; i--) {
    auto [a, b] = edges[i];
    if (!vis[a] && !vis[b]) {
      to[a].erase(b);
      to[b].erase(a);
      nbr[a]--;
      nbr[b]--;
      if (nbr[a] < k) {
        q.push(a);
        vis[a] = true;
        all--;
      }
      if (nbr[b] < k) {
        q.push(b);
        vis[b] = true;
        all--;
      }
      while (!q.empty()) {
        int cur = q.front();
        q.pop();
        for (int t : to[cur]) {
          nbr[t]--;
          to[t].erase(cur);
          if (nbr[t] < k && !vis[t]) {
            q.push(t);
            vis[t] = true;
            all--;
          }
        }
      }
    }
    answer.push_back(all);
  }
  for (int i = answer.size() - 1; i >= 0; i--) {
    cout << answer[i] << "\n";
  }
}

### Prompt

Generate a CPP solution to the following problem:

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 module = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<set<int>> to(n, set<int>());
  vector<pair<int, int>> edges(m);
  vector<int> nbr(n);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    nbr[a]++;
    nbr[b]++;
    to[a].insert(b);
    to[b].insert(a);
    edges[i] = {a, b};
  }
  vector<bool> vis(n);
  int all = n;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      if (nbr[i] < k) {
        queue<int> q;
        q.push(i);
        vis[i] = true;
        all--;
        while (!q.empty()) {
          int cur = q.front();
          q.pop();
          for (int t : to[cur]) {
            nbr[t]--;
            to[t].erase(cur);
            if (nbr[t] < k && !vis[t]) {
              q.push(t);
              vis[t] = true;
              all--;
            }
          }
        }
      }
    }
  }
  queue<int> q;
  vector<int> answer;
  answer.push_back(all);
  for (int i = m - 1; i > 0; i--) {
    auto [a, b] = edges[i];
    if (!vis[a] && !vis[b]) {
      to[a].erase(b);
      to[b].erase(a);
      nbr[a]--;
      nbr[b]--;
      if (nbr[a] < k) {
        q.push(a);
        vis[a] = true;
        all--;
      }
      if (nbr[b] < k) {
        q.push(b);
        vis[b] = true;
        all--;
      }
      while (!q.empty()) {
        int cur = q.front();
        q.pop();
        for (int t : to[cur]) {
          nbr[t]--;
          to[t].erase(cur);
          if (nbr[t] < k && !vis[t]) {
            q.push(t);
            vis[t] = true;
            all--;
          }
        }
      }
    }
    answer.push_back(all);
  }
  for (int i = answer.size() - 1; i >= 0; i--) {
    cout << answer[i] << "\n";
  }
}

```