#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int n;
vector<set<int> > g(200000);
vector<int> used(200000);
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> x(m);
  vector<int> y(m);
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    g[x[i]].insert(y[i]);
    g[y[i]].insert(x[i]);
  }
  set<pair<int, int> > s;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    s.insert(make_pair(g[i].size(), i));
    v[i] = g[i].size();
  }
  while (s.size()) {
    int a = s.begin()->first;
    int b = s.begin()->second;
    if (a < k) {
      s.erase(make_pair(a, b));
      v[b] = 0;
      for (auto i = g[b].begin(); i != g[b].end(); i++)
        if (v[*i] >= k) {
          s.erase(make_pair(v[*i], *i));
          v[*i]--;
          s.insert(make_pair(v[*i], *i));
        }
    } else
      break;
  }
  vector<int> ans(m);
  ans[m - 1] = s.size();
  for (int i = m - 1; i >= 1; i--) {
    int fl = 0;
    if (s.find(make_pair(v[x[i]], x[i])) != s.end()) fl++;
    if (s.find(make_pair(v[y[i]], y[i])) != s.end()) fl++;
    if (fl != 2)
      ans[i - 1] = ans[i];
    else {
      s.erase(make_pair(v[x[i]], x[i]));
      s.erase(make_pair(v[y[i]], y[i]));
      v[x[i]]--;
      v[y[i]]--;
      s.insert(make_pair(v[x[i]], x[i]));
      s.insert(make_pair(v[y[i]], y[i]));
      g[x[i]].erase(y[i]);
      g[y[i]].erase(x[i]);
      while (s.size()) {
        int a = s.begin()->first;
        int b = s.begin()->second;
        if (a < k) {
          s.erase(make_pair(a, b));
          v[b] = 0;
          for (auto i = g[b].begin(); i != g[b].end(); i++)
            if (v[*i] >= k) {
              s.erase(make_pair(v[*i], *i));
              v[*i]--;
              s.insert(make_pair(v[*i], *i));
            }
        } else
          break;
      }
      ans[i - 1] = s.size();
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma warning(disable : 4996)
using namespace std;
int n;
vector<set<int> > g(200000);
vector<int> used(200000);
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> x(m);
  vector<int> y(m);
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    g[x[i]].insert(y[i]);
    g[y[i]].insert(x[i]);
  }
  set<pair<int, int> > s;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    s.insert(make_pair(g[i].size(), i));
    v[i] = g[i].size();
  }
  while (s.size()) {
    int a = s.begin()->first;
    int b = s.begin()->second;
    if (a < k) {
      s.erase(make_pair(a, b));
      v[b] = 0;
      for (auto i = g[b].begin(); i != g[b].end(); i++)
        if (v[*i] >= k) {
          s.erase(make_pair(v[*i], *i));
          v[*i]--;
          s.insert(make_pair(v[*i], *i));
        }
    } else
      break;
  }
  vector<int> ans(m);
  ans[m - 1] = s.size();
  for (int i = m - 1; i >= 1; i--) {
    int fl = 0;
    if (s.find(make_pair(v[x[i]], x[i])) != s.end()) fl++;
    if (s.find(make_pair(v[y[i]], y[i])) != s.end()) fl++;
    if (fl != 2)
      ans[i - 1] = ans[i];
    else {
      s.erase(make_pair(v[x[i]], x[i]));
      s.erase(make_pair(v[y[i]], y[i]));
      v[x[i]]--;
      v[y[i]]--;
      s.insert(make_pair(v[x[i]], x[i]));
      s.insert(make_pair(v[y[i]], y[i]));
      g[x[i]].erase(y[i]);
      g[y[i]].erase(x[i]);
      while (s.size()) {
        int a = s.begin()->first;
        int b = s.begin()->second;
        if (a < k) {
          s.erase(make_pair(a, b));
          v[b] = 0;
          for (auto i = g[b].begin(); i != g[b].end(); i++)
            if (v[*i] >= k) {
              s.erase(make_pair(v[*i], *i));
              v[*i]--;
              s.insert(make_pair(v[*i], *i));
            }
        } else
          break;
      }
      ans[i - 1] = s.size();
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}

```