#include <bits/stdc++.h>
using namespace std;
int n, k;
int res;
vector<set<int>> g;
vector<int> used;
vector<int> selected;
void bfs(int start) {
  vector<int> seen = {start};
  queue<int> q;
  used[start] = 1;
  q.push(start);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    assert(selected[v]);
    assert((int)g[v].size() < k);
    selected[v] = 0;
    res--;
    for (auto t : g[v]) {
      g[t].erase(v);
      if (selected[t] && (int)g[t].size() < k && !used[t]) {
        seen.push_back(t);
        used[t] = 1;
        q.push(t);
      }
    }
  }
  for (auto t : seen) {
    used[t] = 0;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int m;
  cin >> n >> m >> k;
  used.resize(n);
  g.resize(n);
  vector<pair<int, int>> q;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].insert(v);
    g[v].insert(u);
    q.push_back({u, v});
  }
  reverse(q.begin(), q.end());
  selected.assign(n, 1);
  res = n;
  for (int i = 0; i < n; i++) {
    if ((int)g[i].size() < k && selected[i]) {
      bfs(i);
    }
  }
  vector<int> ans;
  for (auto [u, v] : q) {
    ans.push_back(res);
    g[u].erase(v);
    g[v].erase(u);
    if ((int)g[u].size() < k && selected[u]) {
      bfs(u);
    }
    if ((int)g[v].size() < k && selected[v]) {
      bfs(v);
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto t : ans) {
    cout << t << '\n';
  }
}

### 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;
int n, k;
int res;
vector<set<int>> g;
vector<int> used;
vector<int> selected;
void bfs(int start) {
  vector<int> seen = {start};
  queue<int> q;
  used[start] = 1;
  q.push(start);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    assert(selected[v]);
    assert((int)g[v].size() < k);
    selected[v] = 0;
    res--;
    for (auto t : g[v]) {
      g[t].erase(v);
      if (selected[t] && (int)g[t].size() < k && !used[t]) {
        seen.push_back(t);
        used[t] = 1;
        q.push(t);
      }
    }
  }
  for (auto t : seen) {
    used[t] = 0;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int m;
  cin >> n >> m >> k;
  used.resize(n);
  g.resize(n);
  vector<pair<int, int>> q;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].insert(v);
    g[v].insert(u);
    q.push_back({u, v});
  }
  reverse(q.begin(), q.end());
  selected.assign(n, 1);
  res = n;
  for (int i = 0; i < n; i++) {
    if ((int)g[i].size() < k && selected[i]) {
      bfs(i);
    }
  }
  vector<int> ans;
  for (auto [u, v] : q) {
    ans.push_back(res);
    g[u].erase(v);
    g[v].erase(u);
    if ((int)g[u].size() < k && selected[u]) {
      bfs(u);
    }
    if ((int)g[v].size() < k && selected[v]) {
      bfs(v);
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto t : ans) {
    cout << t << '\n';
  }
}

```