#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 3e5 + 10;
long long x[N], y[N], deg[N];
vector<long long> vp[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  for (long long i = 1; i <= m; i++) {
    cin >> x[i] >> y[i];
    vp[x[i]].push_back(y[i]);
    vp[y[i]].push_back(x[i]);
    deg[x[i]]++;
    deg[y[i]]++;
  }
  set<pair<long long, long long> > s;
  for (long long i = 1; i <= n; i++) {
    s.insert({deg[i], i});
  }
  long long ans = n;
  set<pair<long long, long long> > ae;
  vector<long long> v;
  for (long long i = m; i >= 1; i--) {
    while (s.size() > 0) {
      auto it = *s.begin();
      s.erase(it);
      if (it.first >= k) break;
      for (auto it1 : vp[it.second]) {
        if (ae.find({it1, it.second}) == ae.end()) {
          ae.insert({it1, it.second});
          ae.insert({it.second, it1});
          s.erase({deg[it1], it1});
          deg[it1]--;
          s.insert({deg[it1], it1});
        }
      }
      ans--;
    }
    if (ae.find({x[i], y[i]}) == ae.end()) {
      ae.insert({x[i], y[i]});
      ae.insert({y[i], x[i]});
      s.erase({deg[x[i]], x[i]});
      s.erase({deg[y[i]], y[i]});
      deg[x[i]]--;
      s.insert({deg[x[i]], x[i]});
      deg[y[i]]--;
      s.insert({deg[y[i]], y[i]});
    }
    v.push_back(ans);
  }
  reverse(v.begin(), v.end());
  for (auto it : v) {
    cout << it << "\n";
  }
}

### 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>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 3e5 + 10;
long long x[N], y[N], deg[N];
vector<long long> vp[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  for (long long i = 1; i <= m; i++) {
    cin >> x[i] >> y[i];
    vp[x[i]].push_back(y[i]);
    vp[y[i]].push_back(x[i]);
    deg[x[i]]++;
    deg[y[i]]++;
  }
  set<pair<long long, long long> > s;
  for (long long i = 1; i <= n; i++) {
    s.insert({deg[i], i});
  }
  long long ans = n;
  set<pair<long long, long long> > ae;
  vector<long long> v;
  for (long long i = m; i >= 1; i--) {
    while (s.size() > 0) {
      auto it = *s.begin();
      s.erase(it);
      if (it.first >= k) break;
      for (auto it1 : vp[it.second]) {
        if (ae.find({it1, it.second}) == ae.end()) {
          ae.insert({it1, it.second});
          ae.insert({it.second, it1});
          s.erase({deg[it1], it1});
          deg[it1]--;
          s.insert({deg[it1], it1});
        }
      }
      ans--;
    }
    if (ae.find({x[i], y[i]}) == ae.end()) {
      ae.insert({x[i], y[i]});
      ae.insert({y[i], x[i]});
      s.erase({deg[x[i]], x[i]});
      s.erase({deg[y[i]], y[i]});
      deg[x[i]]--;
      s.insert({deg[x[i]], x[i]});
      deg[y[i]]--;
      s.insert({deg[y[i]], y[i]});
    }
    v.push_back(ans);
  }
  reverse(v.begin(), v.end());
  for (auto it : v) {
    cout << it << "\n";
  }
}

```