#include <bits/stdc++.h>
using namespace std;
set<int> v[200005];
int x[200005];
int y[200005];
int an[200005];
set<pair<int, int> > s;
void r(int i, int k) {
  if (v[x[i]].size() >= k && v[y[i]].size() >= k) {
    s.erase(pair<int, int>(v[x[i]].size(), x[i]));
    s.erase(pair<int, int>(v[y[i]].size(), y[i]));
    v[x[i]].erase(y[i]);
    v[y[i]].erase(x[i]);
    s.insert(pair<int, int>(v[x[i]].size(), x[i]));
    s.insert(pair<int, int>(v[y[i]].size(), y[i]));
  }
}
void solve(int k) {
  if (s.empty() || s.begin()->first >= k) return;
  int x = s.begin()->second;
  s.erase(s.begin());
  for (auto it = v[x].begin(); it != v[x].end(); it++)
    if (v[*it].size() >= k) {
      s.erase(pair<int, int>(v[*it].size(), *it));
      v[*it].erase(x);
      s.insert(pair<int, int>(v[*it].size(), *it));
    }
  solve(k);
}
int main() {
  ios::sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    v[x[i]].insert(y[i]);
    v[y[i]].insert(x[i]);
  }
  for (int i = 0; i < n; i++) {
    s.insert(pair<int, int>(v[i + 1].size(), i + 1));
  }
  for (int i = m - 1; i >= 0; i--) {
    solve(k);
    an[i] = s.size();
    r(i, k);
  }
  for (int i = 0; i < m; i++) cout << an[i] << endl;
}

### Prompt

Develop a solution in CPP 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;
set<int> v[200005];
int x[200005];
int y[200005];
int an[200005];
set<pair<int, int> > s;
void r(int i, int k) {
  if (v[x[i]].size() >= k && v[y[i]].size() >= k) {
    s.erase(pair<int, int>(v[x[i]].size(), x[i]));
    s.erase(pair<int, int>(v[y[i]].size(), y[i]));
    v[x[i]].erase(y[i]);
    v[y[i]].erase(x[i]);
    s.insert(pair<int, int>(v[x[i]].size(), x[i]));
    s.insert(pair<int, int>(v[y[i]].size(), y[i]));
  }
}
void solve(int k) {
  if (s.empty() || s.begin()->first >= k) return;
  int x = s.begin()->second;
  s.erase(s.begin());
  for (auto it = v[x].begin(); it != v[x].end(); it++)
    if (v[*it].size() >= k) {
      s.erase(pair<int, int>(v[*it].size(), *it));
      v[*it].erase(x);
      s.insert(pair<int, int>(v[*it].size(), *it));
    }
  solve(k);
}
int main() {
  ios::sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    v[x[i]].insert(y[i]);
    v[y[i]].insert(x[i]);
  }
  for (int i = 0; i < n; i++) {
    s.insert(pair<int, int>(v[i + 1].size(), i + 1));
  }
  for (int i = m - 1; i >= 0; i--) {
    solve(k);
    an[i] = s.size();
    r(i, k);
  }
  for (int i = 0; i < m; i++) cout << an[i] << endl;
}

```