#include <bits/stdc++.h>
using namespace std;
vector<set<int> > g;
int n, m, k;
vector<pair<int, int> > q;
set<int> s;
void remove(int v) {
  if (g[v].size() < k && s.erase(v)) {
    for (auto to : g[v]) g[to].erase(v), remove(to);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> k;
  q.resize(m);
  g.resize(n);
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    g[x].insert(y), g[y].insert(x);
    q.push_back(make_pair(x, y));
  }
  for (int i = 0; i < n; ++i) s.insert(i);
  for (int i = 0; i < n; ++i) remove(i);
  vector<int> ans;
  for (int i = 0; i < m; ++i) {
    ans.push_back(s.size());
    auto cur = q.back();
    q.pop_back();
    g[cur.first].erase(cur.second);
    g[cur.second].erase(cur.first);
    remove(cur.first);
    remove(cur.second);
  }
  for (int i = 0; i < m; ++i) cout << ans[m - i - 1] << '\n';
  return 0;
}

### 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;
vector<set<int> > g;
int n, m, k;
vector<pair<int, int> > q;
set<int> s;
void remove(int v) {
  if (g[v].size() < k && s.erase(v)) {
    for (auto to : g[v]) g[to].erase(v), remove(to);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> k;
  q.resize(m);
  g.resize(n);
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    g[x].insert(y), g[y].insert(x);
    q.push_back(make_pair(x, y));
  }
  for (int i = 0; i < n; ++i) s.insert(i);
  for (int i = 0; i < n; ++i) remove(i);
  vector<int> ans;
  for (int i = 0; i < m; ++i) {
    ans.push_back(s.size());
    auto cur = q.back();
    q.pop_back();
    g[cur.first].erase(cur.second);
    g[cur.second].erase(cur.first);
    remove(cur.first);
    remove(cur.second);
  }
  for (int i = 0; i < m; ++i) cout << ans[m - i - 1] << '\n';
  return 0;
}

```