#include <bits/stdc++.h>
using namespace std;
set<int> adj[200009];
set<int>::iterator it;
int vis[200009];
pair<int, int> edg[200009];
vector<int> bad;
int ses[200009];
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &edg[i].first, &edg[i].second);
    adj[edg[i].first].insert(edg[i].second);
    adj[edg[i].second].insert(edg[i].first);
  }
  int ans = n;
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() < k) {
      vis[i] = 1;
      ans--;
      bad.push_back(i);
    }
  }
  for (int i = m; i >= 1; i--) {
    while (bad.size() > 0) {
      int v = bad.back();
      bad.pop_back();
      for (it = adj[v].begin(); it != adj[v].end(); it++) {
        int u = *it;
        adj[u].erase(v);
        if (vis[u] == 0 && adj[u].size() < k) {
          bad.push_back(u);
          vis[u] = 1;
          ans--;
        }
      }
      adj[v].clear();
    }
    ses[i] = ans;
    int u = edg[i].first;
    int v = edg[i].second;
    if (vis[u] == 1 || vis[v] == 1) continue;
    adj[u].erase(v);
    adj[v].erase(u);
    if (adj[u].size() < k) {
      bad.push_back(u);
      ans--;
      vis[u] = 1;
    }
    if (adj[v].size() < k) {
      bad.push_back(v);
      ans--;
      vis[v] = 1;
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ses[i]);
  }
  return 0;
}

### Prompt

Please formulate 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;
set<int> adj[200009];
set<int>::iterator it;
int vis[200009];
pair<int, int> edg[200009];
vector<int> bad;
int ses[200009];
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &edg[i].first, &edg[i].second);
    adj[edg[i].first].insert(edg[i].second);
    adj[edg[i].second].insert(edg[i].first);
  }
  int ans = n;
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() < k) {
      vis[i] = 1;
      ans--;
      bad.push_back(i);
    }
  }
  for (int i = m; i >= 1; i--) {
    while (bad.size() > 0) {
      int v = bad.back();
      bad.pop_back();
      for (it = adj[v].begin(); it != adj[v].end(); it++) {
        int u = *it;
        adj[u].erase(v);
        if (vis[u] == 0 && adj[u].size() < k) {
          bad.push_back(u);
          vis[u] = 1;
          ans--;
        }
      }
      adj[v].clear();
    }
    ses[i] = ans;
    int u = edg[i].first;
    int v = edg[i].second;
    if (vis[u] == 1 || vis[v] == 1) continue;
    adj[u].erase(v);
    adj[v].erase(u);
    if (adj[u].size() < k) {
      bad.push_back(u);
      ans--;
      vis[u] = 1;
    }
    if (adj[v].size() < k) {
      bad.push_back(v);
      ans--;
      vis[v] = 1;
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ses[i]);
  }
  return 0;
}

```