#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> g(n);
  vector<int> from(m), to(m);
  for (int i = 0; i < m; i++) {
    cin >> from[i] >> to[i];
    from[i]--;
    to[i]--;
    g[from[i]].emplace_back(to[i], i);
    g[to[i]].emplace_back(from[i], i);
  }
  vector<int> res(m), alive(m, 1), que, deg(n);
  for (int i = 0; i < n; i++) {
    deg[i] = (int)g[i].size();
    if (deg[i] < k) que.push_back(i);
  }
  int ind = 0;
  for (int i = m - 1; i >= 0; i--) {
    while (ind < (int)que.size()) {
      int u = que[ind];
      for (auto p : g[u]) {
        if (!alive[p.second]) continue;
        alive[p.second] = 0;
        deg[u]--;
        if (deg[p.first] == k) que.push_back(p.first);
        deg[p.first]--;
      }
      ind++;
    }
    res[i] = n - que.size();
    if (alive[i]) {
      alive[i] = 0;
      if (deg[from[i]] == k) que.push_back(from[i]);
      if (deg[to[i]] == k) que.push_back(to[i]);
      deg[from[i]]--;
      deg[to[i]]--;
    }
  }
  for (int i : res) cout << i << '\n';
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> g(n);
  vector<int> from(m), to(m);
  for (int i = 0; i < m; i++) {
    cin >> from[i] >> to[i];
    from[i]--;
    to[i]--;
    g[from[i]].emplace_back(to[i], i);
    g[to[i]].emplace_back(from[i], i);
  }
  vector<int> res(m), alive(m, 1), que, deg(n);
  for (int i = 0; i < n; i++) {
    deg[i] = (int)g[i].size();
    if (deg[i] < k) que.push_back(i);
  }
  int ind = 0;
  for (int i = m - 1; i >= 0; i--) {
    while (ind < (int)que.size()) {
      int u = que[ind];
      for (auto p : g[u]) {
        if (!alive[p.second]) continue;
        alive[p.second] = 0;
        deg[u]--;
        if (deg[p.first] == k) que.push_back(p.first);
        deg[p.first]--;
      }
      ind++;
    }
    res[i] = n - que.size();
    if (alive[i]) {
      alive[i] = 0;
      if (deg[from[i]] == k) que.push_back(from[i]);
      if (deg[to[i]] == k) que.push_back(to[i]);
      deg[from[i]]--;
      deg[to[i]]--;
    }
  }
  for (int i : res) cout << i << '\n';
}

```