#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<long long, long long> > > friends(n);
  vector<long long> pointers(n);
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    friends[a - 1].push_back(make_pair(b - 1, i));
    friends[b - 1].push_back(make_pair(a - 1, i));
  }
  queue<long long> q;
  set<pair<long long, long long> > good;
  for (long long i = 0; i < n; i++) {
    if (friends[i].size() >= k) {
      good.insert(make_pair(friends[i][k - 1].second, i));
      pointers[i] = k - 1;
    } else {
      pointers[i] = -1;
      q.push(i);
    }
  }
  long long u = m;
  vector<long long> ans(m);
  fill(ans.begin(), ans.end(), 0);
  vector<bool> used(n);
  fill(used.begin(), used.end(), false);
  while (good.size()) {
    while (q.size()) {
      long long V = q.front();
      q.pop();
      used[V] = true;
      for (long long i = 0; i < friends[V].size(); i++) {
        long long to = friends[V][i].first, tm = friends[V][i].second;
        if (pointers[to] == -1) continue;
        long long T = friends[to][pointers[to]].second;
        if (tm > T) continue;
        pointers[to]++;
        while (pointers[to] < friends[to].size()) {
          long long K = friends[to][pointers[to]].first;
          if (used[K] || friends[to][pointers[to]].second >= u)
            pointers[to]++;
          else
            break;
        }
        if (pointers[to] >= friends[to].size()) {
          good.erase(good.find(make_pair(T, to)));
          pointers[to] = -1;
          q.push(to);
        } else {
          good.erase(good.find(make_pair(T, to)));
          T = friends[to][pointers[to]].second;
          good.insert(make_pair(T, to));
        }
      }
    }
    long long S = good.size();
    if (S == 0) break;
    set<pair<long long, long long> >::iterator it = good.end();
    it--;
    pair<long long, long long> P = *it;
    long long tt = P.first;
    for (long long i = tt; i < u; i++) ans[i] = S;
    u = tt;
    while (good.size()) {
      set<pair<long long, long long> >::iterator it = good.end();
      it--;
      pair<long long, long long> P = *it;
      long long ttt = P.first;
      if (tt != ttt) {
        break;
      }
      good.erase(it);
      pointers[P.second] = -1;
      q.push(P.second);
    }
  }
  for (long long i = 0; i < m; i++) cout << ans[i] << "\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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<long long, long long> > > friends(n);
  vector<long long> pointers(n);
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    friends[a - 1].push_back(make_pair(b - 1, i));
    friends[b - 1].push_back(make_pair(a - 1, i));
  }
  queue<long long> q;
  set<pair<long long, long long> > good;
  for (long long i = 0; i < n; i++) {
    if (friends[i].size() >= k) {
      good.insert(make_pair(friends[i][k - 1].second, i));
      pointers[i] = k - 1;
    } else {
      pointers[i] = -1;
      q.push(i);
    }
  }
  long long u = m;
  vector<long long> ans(m);
  fill(ans.begin(), ans.end(), 0);
  vector<bool> used(n);
  fill(used.begin(), used.end(), false);
  while (good.size()) {
    while (q.size()) {
      long long V = q.front();
      q.pop();
      used[V] = true;
      for (long long i = 0; i < friends[V].size(); i++) {
        long long to = friends[V][i].first, tm = friends[V][i].second;
        if (pointers[to] == -1) continue;
        long long T = friends[to][pointers[to]].second;
        if (tm > T) continue;
        pointers[to]++;
        while (pointers[to] < friends[to].size()) {
          long long K = friends[to][pointers[to]].first;
          if (used[K] || friends[to][pointers[to]].second >= u)
            pointers[to]++;
          else
            break;
        }
        if (pointers[to] >= friends[to].size()) {
          good.erase(good.find(make_pair(T, to)));
          pointers[to] = -1;
          q.push(to);
        } else {
          good.erase(good.find(make_pair(T, to)));
          T = friends[to][pointers[to]].second;
          good.insert(make_pair(T, to));
        }
      }
    }
    long long S = good.size();
    if (S == 0) break;
    set<pair<long long, long long> >::iterator it = good.end();
    it--;
    pair<long long, long long> P = *it;
    long long tt = P.first;
    for (long long i = tt; i < u; i++) ans[i] = S;
    u = tt;
    while (good.size()) {
      set<pair<long long, long long> >::iterator it = good.end();
      it--;
      pair<long long, long long> P = *it;
      long long ttt = P.first;
      if (tt != ttt) {
        break;
      }
      good.erase(it);
      pointers[P.second] = -1;
      q.push(P.second);
    }
  }
  for (long long i = 0; i < m; i++) cout << ans[i] << "\n";
}

```