#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const int N = 2e5 + 5;
int n, m, k;
vector<pair<int, int>> edge;
set<int> adj[N];
bool gone[N];
vector<int> res;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    edge.emplace_back(x, y);
    adj[x].emplace(y);
    adj[y].emplace(x);
  }
  set<pair<int, int>> deg;
  for (int i = 1; i <= n; i++) {
    deg.emplace((int)adj[i].size(), i);
  }
  for (int i = m - 1; i >= 0; i--) {
    while (!deg.empty() && ((deg.begin())->first < k)) {
      auto cur = *deg.begin();
      deg.erase(deg.begin());
      int del = cur.second;
      gone[del] = true;
      for (auto j : adj[del]) {
        if (gone[j]) continue;
        deg.erase(make_pair((int)adj[j].size(), j));
        adj[j].erase(del);
        deg.emplace((int)adj[j].size(), j);
      }
    }
    res.emplace_back((int)deg.size());
    auto now = edge[i];
    int a = now.first, b = now.second;
    if (!gone[a]) {
      deg.erase(make_pair((int)adj[a].size(), a));
      adj[a].erase(b);
      deg.emplace((int)adj[a].size(), a);
    }
    if (!gone[b]) {
      deg.erase(make_pair((int)adj[b].size(), b));
      adj[b].erase(a);
      deg.emplace((int)adj[b].size(), b);
    }
  }
  reverse(res.begin(), res.end());
  for (auto i : res) {
    cout << i << '\n';
  }
  return 0;
}

### Prompt

Please formulate 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;
using ll = long long;
using ld = long double;
const int N = 2e5 + 5;
int n, m, k;
vector<pair<int, int>> edge;
set<int> adj[N];
bool gone[N];
vector<int> res;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    edge.emplace_back(x, y);
    adj[x].emplace(y);
    adj[y].emplace(x);
  }
  set<pair<int, int>> deg;
  for (int i = 1; i <= n; i++) {
    deg.emplace((int)adj[i].size(), i);
  }
  for (int i = m - 1; i >= 0; i--) {
    while (!deg.empty() && ((deg.begin())->first < k)) {
      auto cur = *deg.begin();
      deg.erase(deg.begin());
      int del = cur.second;
      gone[del] = true;
      for (auto j : adj[del]) {
        if (gone[j]) continue;
        deg.erase(make_pair((int)adj[j].size(), j));
        adj[j].erase(del);
        deg.emplace((int)adj[j].size(), j);
      }
    }
    res.emplace_back((int)deg.size());
    auto now = edge[i];
    int a = now.first, b = now.second;
    if (!gone[a]) {
      deg.erase(make_pair((int)adj[a].size(), a));
      adj[a].erase(b);
      deg.emplace((int)adj[a].size(), a);
    }
    if (!gone[b]) {
      deg.erase(make_pair((int)adj[b].size(), b));
      adj[b].erase(a);
      deg.emplace((int)adj[b].size(), b);
    }
  }
  reverse(res.begin(), res.end());
  for (auto i : res) {
    cout << i << '\n';
  }
  return 0;
}

```