#include <bits/stdc++.h>
using namespace std;
int u[200005], v[200005];
int cnt[200005];
unordered_set<int> adj[200005];
unordered_set<int> take;
stack<int> ans;
queue<int> cac, toerase;
int n, m, k;
void emptycac() {
  while (cac.size()) {
    int node = cac.front();
    cac.pop();
    for (auto j : adj[node]) {
      if (take.count(j)) {
        adj[j].erase(node);
        if (adj[j].size() < k) {
          take.erase(j);
          cac.push(j);
        }
      }
    }
    while (toerase.size()) {
      take.erase(toerase.front());
      toerase.pop();
    }
  }
}
signed main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> u[i] >> v[i];
    adj[u[i]].insert(v[i]);
    adj[v[i]].insert(u[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() >= k) {
      take.insert(i);
      for (auto j : adj[i]) {
        take.insert(j);
      }
    }
  }
  for (auto i : take) {
    if (adj[i].size() < k) {
      cac.push(i);
      toerase.push(i);
    }
  }
  while (toerase.size()) {
    take.erase(toerase.front());
    toerase.pop();
  }
  emptycac();
  for (int i = m; i >= 1; i--) {
    ans.push(take.size());
    if (take.count(u[i]) && take.count(v[i])) {
      adj[u[i]].erase(v[i]);
      adj[v[i]].erase(u[i]);
      if (adj[v[i]].size() < k) {
        take.erase(v[i]);
        cac.push(v[i]);
      }
      if (adj[u[i]].size() < k) {
        take.erase(u[i]);
        cac.push(u[i]);
      }
      emptycac();
    }
  }
  while (ans.size()) {
    cout << ans.top() << endl;
    ans.pop();
  }
}

### Prompt

In CPP, your task is to solve 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;
int u[200005], v[200005];
int cnt[200005];
unordered_set<int> adj[200005];
unordered_set<int> take;
stack<int> ans;
queue<int> cac, toerase;
int n, m, k;
void emptycac() {
  while (cac.size()) {
    int node = cac.front();
    cac.pop();
    for (auto j : adj[node]) {
      if (take.count(j)) {
        adj[j].erase(node);
        if (adj[j].size() < k) {
          take.erase(j);
          cac.push(j);
        }
      }
    }
    while (toerase.size()) {
      take.erase(toerase.front());
      toerase.pop();
    }
  }
}
signed main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> u[i] >> v[i];
    adj[u[i]].insert(v[i]);
    adj[v[i]].insert(u[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() >= k) {
      take.insert(i);
      for (auto j : adj[i]) {
        take.insert(j);
      }
    }
  }
  for (auto i : take) {
    if (adj[i].size() < k) {
      cac.push(i);
      toerase.push(i);
    }
  }
  while (toerase.size()) {
    take.erase(toerase.front());
    toerase.pop();
  }
  emptycac();
  for (int i = m; i >= 1; i--) {
    ans.push(take.size());
    if (take.count(u[i]) && take.count(v[i])) {
      adj[u[i]].erase(v[i]);
      adj[v[i]].erase(u[i]);
      if (adj[v[i]].size() < k) {
        take.erase(v[i]);
        cac.push(v[i]);
      }
      if (adj[u[i]].size() < k) {
        take.erase(u[i]);
        cac.push(u[i]);
      }
      emptycac();
    }
  }
  while (ans.size()) {
    cout << ans.top() << endl;
    ans.pop();
  }
}

```