#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  int k;
  cin >> k;
  vector<set<int>> g(n);
  vector<int> x(m), y(m);
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i], x[i]--, y[i]--;
    g[x[i]].emplace(y[i]);
    g[y[i]].emplace(x[i]);
  }
  int ans = n;
  queue<int> que;
  vector<int> alive(n, true);
  for (int i = 0; i < n; i++)
    if ((int)g[i].size() < k) {
      que.emplace(i), alive[i] = false, ans--;
    }
  vector<int> res(m);
  for (int i = m - 1; i >= 0; i--) {
    while (!que.empty()) {
      int u = que.front();
      que.pop();
      for (int v : g[u]) g[v].erase(u);
      for (int v : g[u])
        if (alive[v] && (int)g[v].size() < k) {
          que.emplace(v), alive[v] = false, ans--;
        }
      g[u].clear();
    }
    res[i] = ans;
    int u = x[i], v = y[i];
    g[u].erase(v), g[v].erase(u);
    if (alive[u] && (int)g[u].size() < k) {
      que.emplace(u), alive[u] = false, ans--;
    }
    if (alive[v] && (int)g[v].size() < k) {
      que.emplace(v), alive[v] = false, ans--;
    }
  }
  for (int e : res) cout << e << 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;
  cin >> n >> m;
  int k;
  cin >> k;
  vector<set<int>> g(n);
  vector<int> x(m), y(m);
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i], x[i]--, y[i]--;
    g[x[i]].emplace(y[i]);
    g[y[i]].emplace(x[i]);
  }
  int ans = n;
  queue<int> que;
  vector<int> alive(n, true);
  for (int i = 0; i < n; i++)
    if ((int)g[i].size() < k) {
      que.emplace(i), alive[i] = false, ans--;
    }
  vector<int> res(m);
  for (int i = m - 1; i >= 0; i--) {
    while (!que.empty()) {
      int u = que.front();
      que.pop();
      for (int v : g[u]) g[v].erase(u);
      for (int v : g[u])
        if (alive[v] && (int)g[v].size() < k) {
          que.emplace(v), alive[v] = false, ans--;
        }
      g[u].clear();
    }
    res[i] = ans;
    int u = x[i], v = y[i];
    g[u].erase(v), g[v].erase(u);
    if (alive[u] && (int)g[u].size() < k) {
      que.emplace(u), alive[u] = false, ans--;
    }
    if (alive[v] && (int)g[v].size() < k) {
      que.emplace(v), alive[v] = false, ans--;
    }
  }
  for (int e : res) cout << e << endl;
  return 0;
}

```