#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 1;
int n, m, ans[N], cur, done[N], deg[N], k;
vector<pair<int, int> > ve[N];
pair<int, int> e[N];
void bfs(int u, int time) {
  queue<int> qu;
  done[u] = 1;
  cur--;
  qu.push(u);
  int v;
  while (!qu.empty()) {
    u = qu.front();
    qu.pop();
    for (int i = 0; i < ve[u].size(); i++) {
      v = ve[u][i].first;
      if (done[v]) continue;
      if (ve[u][i].second > time) continue;
      deg[u]--;
      deg[v]--;
      if (deg[v] < k) {
        cur--;
        done[v] = 1;
        qu.push(v);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  cur = n;
  int u, v;
  for (int i = 1; i <= m; i++) {
    cin >> e[i].first >> e[i].second;
    u = e[i].first;
    v = e[i].second;
    ve[u].push_back({v, i});
    ve[v].push_back({u, i});
  }
  for (int i = 1; i <= n; i++) deg[i] = ve[i].size();
  for (int i = 1; i <= n; i++) {
    if (deg[i] < k && !done[i]) {
      bfs(i, m);
    }
  }
  ans[m] = cur;
  for (int i = m - 1; i >= 1; i--) {
    u = e[i + 1].first;
    v = e[i + 1].second;
    if (!done[u] && !done[v]) {
      deg[u]--;
      deg[v]--;
      if (deg[u] < k && !done[u]) bfs(u, i);
      if (deg[v] < k && !done[v]) bfs(v, i);
    }
    ans[i] = cur;
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 + 1;
int n, m, ans[N], cur, done[N], deg[N], k;
vector<pair<int, int> > ve[N];
pair<int, int> e[N];
void bfs(int u, int time) {
  queue<int> qu;
  done[u] = 1;
  cur--;
  qu.push(u);
  int v;
  while (!qu.empty()) {
    u = qu.front();
    qu.pop();
    for (int i = 0; i < ve[u].size(); i++) {
      v = ve[u][i].first;
      if (done[v]) continue;
      if (ve[u][i].second > time) continue;
      deg[u]--;
      deg[v]--;
      if (deg[v] < k) {
        cur--;
        done[v] = 1;
        qu.push(v);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  cur = n;
  int u, v;
  for (int i = 1; i <= m; i++) {
    cin >> e[i].first >> e[i].second;
    u = e[i].first;
    v = e[i].second;
    ve[u].push_back({v, i});
    ve[v].push_back({u, i});
  }
  for (int i = 1; i <= n; i++) deg[i] = ve[i].size();
  for (int i = 1; i <= n; i++) {
    if (deg[i] < k && !done[i]) {
      bfs(i, m);
    }
  }
  ans[m] = cur;
  for (int i = m - 1; i >= 1; i--) {
    u = e[i + 1].first;
    v = e[i + 1].second;
    if (!done[u] && !done[v]) {
      deg[u]--;
      deg[v]--;
      if (deg[u] < k && !done[u]) bfs(u, i);
      if (deg[v] < k && !done[v]) bfs(v, i);
    }
    ans[i] = cur;
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
}

```