#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
vector<int> adj[N];
int deg[N], ans[N];
pair<int, int> ed[N];
queue<int> lazy;
bool rem_e[N], rem_v[N];
int k, tot;
void erase(int i) {
  if (rem_e[i]) return;
  rem_e[i] = true;
  for (int u : {ed[i].first, ed[i].second}) {
    if (deg[u] == k) {
      lazy.push(u);
    }
    deg[u]--;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m >> k;
  tot = n;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    ed[i] = {u, v};
    adj[u].push_back(i);
    deg[u]++;
    adj[v].push_back(i);
    deg[v]++;
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) lazy.push(i);
  for (int i = m - 1; i >= 0; i--) {
    while (!lazy.empty()) {
      int u = lazy.front();
      lazy.pop();
      if (rem_v[u]) continue;
      rem_v[u] = true;
      tot--;
      for (int i : adj[u]) erase(i);
    }
    ans[i] = tot;
    erase(i);
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << ' ';
  }
  cout << '\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;
const int N = 2e5 + 10;
vector<int> adj[N];
int deg[N], ans[N];
pair<int, int> ed[N];
queue<int> lazy;
bool rem_e[N], rem_v[N];
int k, tot;
void erase(int i) {
  if (rem_e[i]) return;
  rem_e[i] = true;
  for (int u : {ed[i].first, ed[i].second}) {
    if (deg[u] == k) {
      lazy.push(u);
    }
    deg[u]--;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m >> k;
  tot = n;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    ed[i] = {u, v};
    adj[u].push_back(i);
    deg[u]++;
    adj[v].push_back(i);
    deg[v]++;
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) lazy.push(i);
  for (int i = m - 1; i >= 0; i--) {
    while (!lazy.empty()) {
      int u = lazy.front();
      lazy.pop();
      if (rem_v[u]) continue;
      rem_v[u] = true;
      tot--;
      for (int i : adj[u]) erase(i);
    }
    ans[i] = tot;
    erase(i);
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
  return 0;
}

```