#include <bits/stdc++.h>
using namespace std;
const int INFTY = 20000000;
const int MAX = 500100;
const int MOD = 10000000;
void coutTab(int* tab, int n) {
  for (int i = 0; i < n; i++) {
    cout << tab[i] << " ";
  }
  cout << "\n";
}
int n, m, k;
vector<set<int> > G(MAX);
int s[MAX];
bool was[MAX];
pair<int, int> E[MAX];
int res[MAX];
void addToQueue(queue<int>& Q, int i) {
  if (s[i] < k && !was[i]) {
    was[i] = 1;
    Q.push(i);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> E[i].first >> E[i].second;
    E[i].first--;
    E[i].second--;
    G[E[i].first].insert(E[i].second);
    G[E[i].second].insert(E[i].first);
  }
  for (int i = 0; i < n; i++) s[i] = (G[i]).size();
  queue<int> Q;
  int cur = n;
  for (int i = 0; i < n; i++) addToQueue(Q, i);
  for (int i = 0; i < m; i++) {
    while (!Q.empty()) {
      int j = Q.front();
      Q.pop();
      for (auto l : G[j]) {
        s[l]--;
        addToQueue(Q, l);
      }
      cur--;
    }
    res[i] = cur;
    pair<int, int> lst = E[m - i - 1];
    G[lst.first].erase(lst.second);
    G[lst.second].erase(lst.first);
    if (!was[lst.second]) s[lst.first]--;
    if (!was[lst.first]) s[lst.second]--;
    addToQueue(Q, lst.first);
    addToQueue(Q, lst.second);
  }
  reverse(res, res + m);
  for (int i = 0; i < m; i++) cout << res[i] << "\n";
}

### Prompt

Create a solution in Cpp for 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 INFTY = 20000000;
const int MAX = 500100;
const int MOD = 10000000;
void coutTab(int* tab, int n) {
  for (int i = 0; i < n; i++) {
    cout << tab[i] << " ";
  }
  cout << "\n";
}
int n, m, k;
vector<set<int> > G(MAX);
int s[MAX];
bool was[MAX];
pair<int, int> E[MAX];
int res[MAX];
void addToQueue(queue<int>& Q, int i) {
  if (s[i] < k && !was[i]) {
    was[i] = 1;
    Q.push(i);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> E[i].first >> E[i].second;
    E[i].first--;
    E[i].second--;
    G[E[i].first].insert(E[i].second);
    G[E[i].second].insert(E[i].first);
  }
  for (int i = 0; i < n; i++) s[i] = (G[i]).size();
  queue<int> Q;
  int cur = n;
  for (int i = 0; i < n; i++) addToQueue(Q, i);
  for (int i = 0; i < m; i++) {
    while (!Q.empty()) {
      int j = Q.front();
      Q.pop();
      for (auto l : G[j]) {
        s[l]--;
        addToQueue(Q, l);
      }
      cur--;
    }
    res[i] = cur;
    pair<int, int> lst = E[m - i - 1];
    G[lst.first].erase(lst.second);
    G[lst.second].erase(lst.first);
    if (!was[lst.second]) s[lst.first]--;
    if (!was[lst.first]) s[lst.second]--;
    addToQueue(Q, lst.first);
    addToQueue(Q, lst.second);
  }
  reverse(res, res + m);
  for (int i = 0; i < m; i++) cout << res[i] << "\n";
}

```