#include <bits/stdc++.h>
using namespace std;
int n, m, d[200005], k, res, vs[200005], ans[200005], pa;
map<int, int> cnt[200005];
vector<int> ke[200005];
pair<int, int> a[200005];
void dfs(int u) {
  d[u]--;
  res--;
  vs[u] = 1;
  for (int v : ke[u])
    if (vs[v] == 0) {
      cnt[u][v] = cnt[v][u] = 1;
      d[v]--;
      if (d[v] < k) {
        d[v]++;
        dfs(v);
      }
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  res = n;
  for (int i = 1; i <= m; i++) {
    cin >> a[i].first >> a[i].second;
    d[a[i].first]++;
    d[a[i].second]++;
    ke[a[i].first].push_back(a[i].second);
    ke[a[i].second].push_back(a[i].first);
  }
  for (int i = 1; i <= n; i++)
    if (d[i] < k && vs[i] == 0) dfs(i);
  for (int i = m; i >= 1; i--) {
    ans[i] = res;
    ke[a[i].first].pop_back();
    ke[a[i].second].pop_back();
    if (cnt[a[i].first][a[i].second] == 1) continue;
    if (vs[a[i].first] == 0) {
      d[a[i].first]--;
      if (d[a[i].first] < k) {
        d[a[i].first]++;
        dfs(a[i].first);
      }
    }
    if (vs[a[i].second] == 0) {
      d[a[i].second]--;
      if (d[a[i].second] < k) {
        d[a[i].second]++;
        dfs(a[i].second);
      }
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
}

### Prompt

Please formulate 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 n, m, d[200005], k, res, vs[200005], ans[200005], pa;
map<int, int> cnt[200005];
vector<int> ke[200005];
pair<int, int> a[200005];
void dfs(int u) {
  d[u]--;
  res--;
  vs[u] = 1;
  for (int v : ke[u])
    if (vs[v] == 0) {
      cnt[u][v] = cnt[v][u] = 1;
      d[v]--;
      if (d[v] < k) {
        d[v]++;
        dfs(v);
      }
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  res = n;
  for (int i = 1; i <= m; i++) {
    cin >> a[i].first >> a[i].second;
    d[a[i].first]++;
    d[a[i].second]++;
    ke[a[i].first].push_back(a[i].second);
    ke[a[i].second].push_back(a[i].first);
  }
  for (int i = 1; i <= n; i++)
    if (d[i] < k && vs[i] == 0) dfs(i);
  for (int i = m; i >= 1; i--) {
    ans[i] = res;
    ke[a[i].first].pop_back();
    ke[a[i].second].pop_back();
    if (cnt[a[i].first][a[i].second] == 1) continue;
    if (vs[a[i].first] == 0) {
      d[a[i].first]--;
      if (d[a[i].first] < k) {
        d[a[i].first]++;
        dfs(a[i].first);
      }
    }
    if (vs[a[i].second] == 0) {
      d[a[i].second]--;
      if (d[a[i].second] < k) {
        d[a[i].second]++;
        dfs(a[i].second);
      }
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
}

```