#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  const int LIM = 600;
  int N, M, K;
  cin >> N >> M >> K;
  vector<int> x(M), y(M);
  vector<vector<int> > g(N);
  for (int i = 0; i < M; i++) {
    cin >> x[i] >> y[i];
    g[--x[i]].push_back(i);
    g[--y[i]].push_back(i);
  }
  vector<int> deg(N);
  for (int i = 0; i < N; i++) deg[i] = g[i].size();
  vector<bool> del(N, false);
  vector<bool> live(M, true);
  int alive = N;
  vector<int> ans(M);
  auto bfs = [&](int v) {
    if (deg[v] >= K || del[v]) return;
    queue<int> rem;
    rem.push(v);
    del[v] = true;
    alive--;
    while (!rem.empty()) {
      int nxt = rem.front();
      rem.pop();
      for (int e : g[nxt]) {
        if (!live[e]) continue;
        live[e] = false;
        int nbr = x[e] + y[e] - nxt;
        deg[nbr]--;
        if (!del[nbr] && deg[nbr] < K) {
          del[nbr] = true;
          alive--;
          rem.push(nbr);
        }
      }
    }
  };
  for (int i = 0; i < N; i++) {
    bfs(i);
  }
  for (int i = M - 1; i >= 0; i--) {
    ans[i] = alive;
    if (live[i]) {
      live[i] = false;
      deg[x[i]]--;
      deg[y[i]]--;
      bfs(x[i]);
      bfs(y[i]);
    }
  }
  for (int i = 0; i < M; i++) cout << ans[i] << "\n";
  return 0;
}

### 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  const int LIM = 600;
  int N, M, K;
  cin >> N >> M >> K;
  vector<int> x(M), y(M);
  vector<vector<int> > g(N);
  for (int i = 0; i < M; i++) {
    cin >> x[i] >> y[i];
    g[--x[i]].push_back(i);
    g[--y[i]].push_back(i);
  }
  vector<int> deg(N);
  for (int i = 0; i < N; i++) deg[i] = g[i].size();
  vector<bool> del(N, false);
  vector<bool> live(M, true);
  int alive = N;
  vector<int> ans(M);
  auto bfs = [&](int v) {
    if (deg[v] >= K || del[v]) return;
    queue<int> rem;
    rem.push(v);
    del[v] = true;
    alive--;
    while (!rem.empty()) {
      int nxt = rem.front();
      rem.pop();
      for (int e : g[nxt]) {
        if (!live[e]) continue;
        live[e] = false;
        int nbr = x[e] + y[e] - nxt;
        deg[nbr]--;
        if (!del[nbr] && deg[nbr] < K) {
          del[nbr] = true;
          alive--;
          rem.push(nbr);
        }
      }
    }
  };
  for (int i = 0; i < N; i++) {
    bfs(i);
  }
  for (int i = M - 1; i >= 0; i--) {
    ans[i] = alive;
    if (live[i]) {
      live[i] = false;
      deg[x[i]]--;
      deg[y[i]]--;
      bfs(x[i]);
      bfs(y[i]);
    }
  }
  for (int i = 0; i < M; i++) cout << ans[i] << "\n";
  return 0;
}

```