#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int m, n, k, d[N];
set<int> ad[N];
int ans[N], dd[N];
pair<int, int> ed[N];
int main() {
  ;
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    ad[u].insert(v);
    ad[v].insert(u);
    d[u]++;
    d[v]++;
    ed[i] = {u, v};
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (d[i] < k) q.push(i), dd[i] = 1;
  ans[m] = n;
  while (!q.empty()) {
    int uu = q.front();
    q.pop();
    for (auto &vv : ad[uu]) {
      d[vv]--;
      ad[vv].erase(uu);
      if (d[vv] < k && !dd[vv]) {
        dd[vv] = 1;
        q.push(vv);
      }
    }
    ans[m]--;
    d[uu] = 0;
    ad[uu].clear();
  }
  for (int i = m - 1; i >= 1; --i) {
    ans[i] = ans[i + 1];
    if (dd[ed[i + 1].first] || dd[ed[i + 1].second]) continue;
    int u = ed[i + 1].first, v = ed[i + 1].second;
    ad[u].erase(v);
    ad[v].erase(u);
    d[u]--;
    d[v]--;
    if (d[u] < k && !dd[u]) q.push(u), dd[u] = 1;
    if (d[v] < k && !dd[v]) q.push(v), dd[v] = 1;
    while (!q.empty()) {
      int uu = q.front();
      q.pop();
      for (auto &vv : ad[uu]) {
        d[vv]--;
        ad[vv].erase(uu);
        if (d[vv] < k && !dd[vv]) {
          dd[vv] = 1;
          q.push(vv);
        }
      }
      ans[i]--;
      d[uu] = 0;
      ad[uu].clear();
    }
  }
  for (int i = 1; i <= m; ++i) cout << ans[i] << '\n';
}

### 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 + 5;
int m, n, k, d[N];
set<int> ad[N];
int ans[N], dd[N];
pair<int, int> ed[N];
int main() {
  ;
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    ad[u].insert(v);
    ad[v].insert(u);
    d[u]++;
    d[v]++;
    ed[i] = {u, v};
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (d[i] < k) q.push(i), dd[i] = 1;
  ans[m] = n;
  while (!q.empty()) {
    int uu = q.front();
    q.pop();
    for (auto &vv : ad[uu]) {
      d[vv]--;
      ad[vv].erase(uu);
      if (d[vv] < k && !dd[vv]) {
        dd[vv] = 1;
        q.push(vv);
      }
    }
    ans[m]--;
    d[uu] = 0;
    ad[uu].clear();
  }
  for (int i = m - 1; i >= 1; --i) {
    ans[i] = ans[i + 1];
    if (dd[ed[i + 1].first] || dd[ed[i + 1].second]) continue;
    int u = ed[i + 1].first, v = ed[i + 1].second;
    ad[u].erase(v);
    ad[v].erase(u);
    d[u]--;
    d[v]--;
    if (d[u] < k && !dd[u]) q.push(u), dd[u] = 1;
    if (d[v] < k && !dd[v]) q.push(v), dd[v] = 1;
    while (!q.empty()) {
      int uu = q.front();
      q.pop();
      for (auto &vv : ad[uu]) {
        d[vv]--;
        ad[vv].erase(uu);
        if (d[vv] < k && !dd[vv]) {
          dd[vv] = 1;
          q.push(vv);
        }
      }
      ans[i]--;
      d[uu] = 0;
      ad[uu].clear();
    }
  }
  for (int i = 1; i <= m; ++i) cout << ans[i] << '\n';
}

```