#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, i, k, u;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  vector<pair<int, int> > adj[n + 5];
  pair<int, int> edges[m + 5];
  set<pair<int, int> > degree_set;
  set<pair<int, int> >::iterator it;
  bool isthere[n + 5];
  int deg[n + 5];
  int ans[m + 5];
  memset(ans, 0, sizeof(ans));
  memset(deg, 0, sizeof(deg));
  memset(isthere, true, sizeof(isthere));
  for (i = 0; i < m; i++) {
    scanf("%d", &edges[i].first);
    scanf("%d", &edges[i].second);
    adj[edges[i].first].push_back({edges[i].second, i});
    adj[edges[i].second].push_back({edges[i].first, i});
    ++deg[edges[i].first];
    degree_set.insert({deg[edges[i].first], edges[i].first});
    degree_set.erase({deg[edges[i].first] - 1, edges[i].first});
    ++deg[edges[i].second];
    degree_set.insert({deg[edges[i].second], edges[i].second});
    degree_set.erase({deg[edges[i].second] - 1, edges[i].second});
  }
  for (i = m - 1; i >= 0; i--) {
    while (!degree_set.empty() && (*(degree_set.begin())).first < k) {
      u = (*(degree_set.begin())).second;
      for (auto k1 : adj[u]) {
        if (isthere[k1.first] == true && k1.second <= i) {
          degree_set.erase({deg[k1.first], k1.first});
          deg[k1.first]--;
          degree_set.insert({deg[k1.first], k1.first});
        }
      }
      degree_set.erase({deg[u], u});
      isthere[u] = false;
    }
    ans[i] = degree_set.size();
    if (isthere[edges[i].first] == true && isthere[edges[i].second] == true) {
      degree_set.erase({deg[edges[i].first], edges[i].first});
      deg[edges[i].first]--;
      degree_set.insert({deg[edges[i].first], edges[i].first});
      degree_set.erase({deg[edges[i].second], edges[i].second});
      deg[edges[i].second]--;
      degree_set.insert({deg[edges[i].second], edges[i].second});
    }
  }
  for (i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### 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 main() {
  int n, m, i, k, u;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  vector<pair<int, int> > adj[n + 5];
  pair<int, int> edges[m + 5];
  set<pair<int, int> > degree_set;
  set<pair<int, int> >::iterator it;
  bool isthere[n + 5];
  int deg[n + 5];
  int ans[m + 5];
  memset(ans, 0, sizeof(ans));
  memset(deg, 0, sizeof(deg));
  memset(isthere, true, sizeof(isthere));
  for (i = 0; i < m; i++) {
    scanf("%d", &edges[i].first);
    scanf("%d", &edges[i].second);
    adj[edges[i].first].push_back({edges[i].second, i});
    adj[edges[i].second].push_back({edges[i].first, i});
    ++deg[edges[i].first];
    degree_set.insert({deg[edges[i].first], edges[i].first});
    degree_set.erase({deg[edges[i].first] - 1, edges[i].first});
    ++deg[edges[i].second];
    degree_set.insert({deg[edges[i].second], edges[i].second});
    degree_set.erase({deg[edges[i].second] - 1, edges[i].second});
  }
  for (i = m - 1; i >= 0; i--) {
    while (!degree_set.empty() && (*(degree_set.begin())).first < k) {
      u = (*(degree_set.begin())).second;
      for (auto k1 : adj[u]) {
        if (isthere[k1.first] == true && k1.second <= i) {
          degree_set.erase({deg[k1.first], k1.first});
          deg[k1.first]--;
          degree_set.insert({deg[k1.first], k1.first});
        }
      }
      degree_set.erase({deg[u], u});
      isthere[u] = false;
    }
    ans[i] = degree_set.size();
    if (isthere[edges[i].first] == true && isthere[edges[i].second] == true) {
      degree_set.erase({deg[edges[i].first], edges[i].first});
      deg[edges[i].first]--;
      degree_set.insert({deg[edges[i].first], edges[i].first});
      degree_set.erase({deg[edges[i].second], edges[i].second});
      deg[edges[i].second]--;
      degree_set.insert({deg[edges[i].second], edges[i].second});
    }
  }
  for (i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```