#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<pair<int, int> > arr(m);
  vector<vector<int> > G(n);
  vector<int> in(n), vis(n);
  for (auto& it : arr) {
    int u, v;
    cin >> u >> v;
    in[--u]++;
    in[--v]++;
    it = {u, v};
    G[u].push_back(v);
    G[v].push_back(u);
  }
  queue<int> q;
  for (int i = 0; i < n; i++) {
    if (in[i] < k) {
      q.push(i);
      vis[i] = 1;
    }
  }
  int cur = n - q.size();
  set<pair<int, int> > cnt;
  while (!q.empty()) {
    auto x = q.front();
    q.pop();
    for (auto& it : G[x]) {
      auto node = minmax(x, it);
      if (cnt.find(node) == cnt.end()) {
        cnt.insert(node);
        if (!vis[it] && --in[it] < k) {
          vis[it] = 1;
          q.push(it);
          cur--;
        }
      }
    }
  }
  reverse(arr.begin(), arr.end());
  vector<int> ans;
  for (auto& it : arr) {
    ans.push_back(cur);
    auto node = minmax(it.first, it.second);
    if (cnt.find(node) == cnt.end()) {
      cnt.insert(node);
      if (!vis[it.first] && --in[it.first] < k) {
        vis[it.first] = 1;
        q.push(it.first);
        cur--;
      }
      if (!vis[it.second] && --in[it.second] < k) {
        vis[it.second] = 1;
        q.push(it.second);
        cur--;
      }
      while (!q.empty()) {
        auto x = q.front();
        q.pop();
        for (auto& jt : G[x]) {
          auto node = minmax(x, jt);
          if (cnt.find(node) == cnt.end()) {
            cnt.insert(node);
            if (!vis[jt] && --in[jt] < k) {
              vis[jt] = 1;
              q.push(jt);
              cur--;
            }
          }
        }
      }
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto& it : ans) cout << it << 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 main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<pair<int, int> > arr(m);
  vector<vector<int> > G(n);
  vector<int> in(n), vis(n);
  for (auto& it : arr) {
    int u, v;
    cin >> u >> v;
    in[--u]++;
    in[--v]++;
    it = {u, v};
    G[u].push_back(v);
    G[v].push_back(u);
  }
  queue<int> q;
  for (int i = 0; i < n; i++) {
    if (in[i] < k) {
      q.push(i);
      vis[i] = 1;
    }
  }
  int cur = n - q.size();
  set<pair<int, int> > cnt;
  while (!q.empty()) {
    auto x = q.front();
    q.pop();
    for (auto& it : G[x]) {
      auto node = minmax(x, it);
      if (cnt.find(node) == cnt.end()) {
        cnt.insert(node);
        if (!vis[it] && --in[it] < k) {
          vis[it] = 1;
          q.push(it);
          cur--;
        }
      }
    }
  }
  reverse(arr.begin(), arr.end());
  vector<int> ans;
  for (auto& it : arr) {
    ans.push_back(cur);
    auto node = minmax(it.first, it.second);
    if (cnt.find(node) == cnt.end()) {
      cnt.insert(node);
      if (!vis[it.first] && --in[it.first] < k) {
        vis[it.first] = 1;
        q.push(it.first);
        cur--;
      }
      if (!vis[it.second] && --in[it.second] < k) {
        vis[it.second] = 1;
        q.push(it.second);
        cur--;
      }
      while (!q.empty()) {
        auto x = q.front();
        q.pop();
        for (auto& jt : G[x]) {
          auto node = minmax(x, jt);
          if (cnt.find(node) == cnt.end()) {
            cnt.insert(node);
            if (!vis[jt] && --in[jt] < k) {
              vis[jt] = 1;
              q.push(jt);
              cur--;
            }
          }
        }
      }
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto& it : ans) cout << it << endl;
  return 0;
}

```