#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, k;
  while (cin >> n >> m >> k) {
    set<pair<int, int> > st;
    vector<vector<int> > G(n + 1);
    vector<int> deg(n + 1), vis(n + 1), x(m), y(m), ans(m);
    for (int i = 0; i < m; i++) {
      cin >> x[i] >> y[i];
      deg[x[i]]++;
      deg[y[i]]++;
      G[x[i]].push_back(y[i]);
      G[y[i]].push_back(x[i]);
    }
    queue<int> q;
    for (int i = 1; i <= n; i++) {
      if (deg[i] < k) {
        vis[i] = true;
        q.push(i);
      }
    }
    for (int cnt = n, day = m - 1; day >= 0; day--) {
      if (day != m - 1) {
        int u = x[day + 1], v = y[day + 1];
        if (!st.count(make_pair(u, v))) {
          st.insert(make_pair(u, v));
          st.insert(make_pair(v, u));
          if (--deg[u] < k && !vis[u]) {
            vis[u] = true;
            q.push(u);
          }
          if (--deg[v] < k && !vis[v]) {
            vis[v] = true;
            q.push(v);
          }
        }
      }
      while (!q.empty()) {
        int u = q.front();
        q.pop();
        cnt--;
        for (int i = 0; i < G[u].size(); i++) {
          int v = G[u][i];
          if (!st.count(make_pair(u, v))) {
            st.insert(make_pair(u, v));
            st.insert(make_pair(v, u));
            if (--deg[v] < k && !vis[v]) {
              vis[v] = true;
              q.push(v);
            }
          }
        }
      }
      ans[day] = cnt;
    }
    for (int i = 0; i < m; i++) {
      cout << ans[i] << endl;
    }
  }
  return 0;
}

### Prompt

Generate 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;
int main() {
  int n, m, k;
  while (cin >> n >> m >> k) {
    set<pair<int, int> > st;
    vector<vector<int> > G(n + 1);
    vector<int> deg(n + 1), vis(n + 1), x(m), y(m), ans(m);
    for (int i = 0; i < m; i++) {
      cin >> x[i] >> y[i];
      deg[x[i]]++;
      deg[y[i]]++;
      G[x[i]].push_back(y[i]);
      G[y[i]].push_back(x[i]);
    }
    queue<int> q;
    for (int i = 1; i <= n; i++) {
      if (deg[i] < k) {
        vis[i] = true;
        q.push(i);
      }
    }
    for (int cnt = n, day = m - 1; day >= 0; day--) {
      if (day != m - 1) {
        int u = x[day + 1], v = y[day + 1];
        if (!st.count(make_pair(u, v))) {
          st.insert(make_pair(u, v));
          st.insert(make_pair(v, u));
          if (--deg[u] < k && !vis[u]) {
            vis[u] = true;
            q.push(u);
          }
          if (--deg[v] < k && !vis[v]) {
            vis[v] = true;
            q.push(v);
          }
        }
      }
      while (!q.empty()) {
        int u = q.front();
        q.pop();
        cnt--;
        for (int i = 0; i < G[u].size(); i++) {
          int v = G[u][i];
          if (!st.count(make_pair(u, v))) {
            st.insert(make_pair(u, v));
            st.insert(make_pair(v, u));
            if (--deg[v] < k && !vis[v]) {
              vis[v] = true;
              q.push(v);
            }
          }
        }
      }
      ans[day] = cnt;
    }
    for (int i = 0; i < m; i++) {
      cout << ans[i] << endl;
    }
  }
  return 0;
}

```