#include <bits/stdc++.h>
using namespace std;
int n, m, k;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  vector<set<int>> g(n);
  vector<int> deg(n);
  set<pair<int, int>> cur;
  vector<pair<int, int>> edges;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    deg[a]++;
    deg[b]++;
    edges.emplace_back(a, b);
    g[a].insert(b);
    g[b].insert(a);
  }
  for (int i = 0; i < n; i++) cur.insert(make_pair(deg[i], i));
  auto solve = [&]() {
    while (true) {
      if (cur.empty()) break;
      auto u = cur.begin();
      vector<int> rev;
      if (u->first < k) {
        for (int i : g[u->second]) {
          rev.push_back(i);
        }
        cur.erase(u);
        for (int i : rev) {
          if (cur.find(make_pair(deg[i], i)) != cur.end()) {
            cur.erase(make_pair(deg[i], i));
            cur.insert(make_pair(deg[i] - 1, i));
          }
          deg[i]--;
        }
      } else
        break;
    }
  };
  solve();
  vector<int> ans;
  ans.push_back(cur.size());
  for (int i = m - 1; i > -1; i--) {
    int u = edges[i].first, v = edges[i].second;
    if (cur.find(make_pair(deg[u], u)) != cur.end() &&
        cur.find(make_pair(deg[v], v)) != cur.end()) {
      cur.erase(make_pair(deg[u], u));
      deg[u]--;
      cur.insert(make_pair(deg[u], u));
      cur.erase(make_pair(deg[v], v));
      deg[v]--;
      cur.insert(make_pair(deg[v], v));
    }
    g[u].erase(v);
    g[v].erase(u);
    solve();
    ans.push_back(cur.size());
  }
  ans.pop_back();
  reverse(ans.begin(), ans.end());
  for (int i : ans) cout << i << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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, m, k;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  vector<set<int>> g(n);
  vector<int> deg(n);
  set<pair<int, int>> cur;
  vector<pair<int, int>> edges;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    deg[a]++;
    deg[b]++;
    edges.emplace_back(a, b);
    g[a].insert(b);
    g[b].insert(a);
  }
  for (int i = 0; i < n; i++) cur.insert(make_pair(deg[i], i));
  auto solve = [&]() {
    while (true) {
      if (cur.empty()) break;
      auto u = cur.begin();
      vector<int> rev;
      if (u->first < k) {
        for (int i : g[u->second]) {
          rev.push_back(i);
        }
        cur.erase(u);
        for (int i : rev) {
          if (cur.find(make_pair(deg[i], i)) != cur.end()) {
            cur.erase(make_pair(deg[i], i));
            cur.insert(make_pair(deg[i] - 1, i));
          }
          deg[i]--;
        }
      } else
        break;
    }
  };
  solve();
  vector<int> ans;
  ans.push_back(cur.size());
  for (int i = m - 1; i > -1; i--) {
    int u = edges[i].first, v = edges[i].second;
    if (cur.find(make_pair(deg[u], u)) != cur.end() &&
        cur.find(make_pair(deg[v], v)) != cur.end()) {
      cur.erase(make_pair(deg[u], u));
      deg[u]--;
      cur.insert(make_pair(deg[u], u));
      cur.erase(make_pair(deg[v], v));
      deg[v]--;
      cur.insert(make_pair(deg[v], v));
    }
    g[u].erase(v);
    g[v].erase(u);
    solve();
    ans.push_back(cur.size());
  }
  ans.pop_back();
  reverse(ans.begin(), ans.end());
  for (int i : ans) cout << i << endl;
  return 0;
}

```