#include <bits/stdc++.h>
using namespace std;
const int mxN = 2e5;
int n, m, k, eu[mxN], ev[mxN], ans[mxN];
set<int> adj[mxN];
bool a[mxN];
vector<int> tp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    cin >> eu[i] >> ev[i], --eu[i], --ev[i];
    adj[eu[i]].insert(i);
    adj[ev[i]].insert(i);
    a[i] = 1;
  }
  for (int i = 0; i < n; ++i)
    if (adj[i].size() < k) tp.push_back(i);
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = i < m - 1 ? ans[i + 1] : n;
    for (int j = 0; j < tp.size(); ++j) {
      for (int e : adj[tp[j]]) {
        adj[eu[e] ^ ev[e] ^ tp[j]].erase(e);
        if (adj[eu[e] ^ ev[e] ^ tp[j]].size() == k - 1)
          tp.push_back(eu[e] ^ ev[e] ^ tp[j]);
        a[e] = 0;
      }
      adj[tp[j]].clear();
      --ans[i];
    }
    tp.clear();
    if (a[i]) {
      adj[eu[i]].erase(i);
      if (adj[eu[i]].size() < k) tp.push_back(eu[i]);
      adj[ev[i]].erase(i);
      if (adj[ev[i]].size() < k) tp.push_back(ev[i]);
      a[i] = 0;
    }
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << "\n";
}

### Prompt

Your task is to create 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;
const int mxN = 2e5;
int n, m, k, eu[mxN], ev[mxN], ans[mxN];
set<int> adj[mxN];
bool a[mxN];
vector<int> tp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    cin >> eu[i] >> ev[i], --eu[i], --ev[i];
    adj[eu[i]].insert(i);
    adj[ev[i]].insert(i);
    a[i] = 1;
  }
  for (int i = 0; i < n; ++i)
    if (adj[i].size() < k) tp.push_back(i);
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = i < m - 1 ? ans[i + 1] : n;
    for (int j = 0; j < tp.size(); ++j) {
      for (int e : adj[tp[j]]) {
        adj[eu[e] ^ ev[e] ^ tp[j]].erase(e);
        if (adj[eu[e] ^ ev[e] ^ tp[j]].size() == k - 1)
          tp.push_back(eu[e] ^ ev[e] ^ tp[j]);
        a[e] = 0;
      }
      adj[tp[j]].clear();
      --ans[i];
    }
    tp.clear();
    if (a[i]) {
      adj[eu[i]].erase(i);
      if (adj[eu[i]].size() < k) tp.push_back(eu[i]);
      adj[ev[i]].erase(i);
      if (adj[ev[i]].size() < k) tp.push_back(ev[i]);
      a[i] = 0;
    }
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << "\n";
}

```