#include <bits/stdc++.h>
using namespace std;
int n, m, k, cnt;
struct Edge {
  int u, v;
};
vector<Edge> vE;
vector<int> ans;
const int maxN = 2 * 100000 + 1;
set<int> edge[maxN];
set<int> nodes;
queue<int> Q;
int do_remove() {
  while (!Q.empty()) {
    int s = Q.front();
    Q.pop();
    if (edge[s].size() < k) {
      for (auto iter1 = edge[s].begin(); iter1 != edge[s].end(); iter1++) {
        int t = *iter1;
        edge[t].erase(s);
        if (edge[t].size() < k) Q.push(t);
      }
      edge[s].clear();
      auto iter = nodes.find(s);
      if (iter != nodes.end()) nodes.erase(iter);
    };
  }
  return nodes.size();
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) nodes.insert(i);
  cnt = n;
  for (int i = 0; i < m; i++) {
    int s, t;
    cin >> s >> t;
    vE.push_back(Edge{s, t});
    edge[s].insert(t);
    edge[t].insert(s);
  }
  for (int i = 1; i <= n; i++)
    if (edge[i].size() < k) Q.push(i);
  int ret = do_remove();
  ans.push_back(ret);
  for (int i = vE.size() - 1; i >= 1; i--) {
    int s = vE[i].u;
    int t = vE[i].v;
    edge[s].erase(t);
    edge[t].erase(s);
    Q.push(s);
    Q.push(t);
    int ret = do_remove();
    ans.push_back(ret);
  }
  for (int i = ans.size() - 1; i >= 0; i--) {
    cout << ans[i] << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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, k, cnt;
struct Edge {
  int u, v;
};
vector<Edge> vE;
vector<int> ans;
const int maxN = 2 * 100000 + 1;
set<int> edge[maxN];
set<int> nodes;
queue<int> Q;
int do_remove() {
  while (!Q.empty()) {
    int s = Q.front();
    Q.pop();
    if (edge[s].size() < k) {
      for (auto iter1 = edge[s].begin(); iter1 != edge[s].end(); iter1++) {
        int t = *iter1;
        edge[t].erase(s);
        if (edge[t].size() < k) Q.push(t);
      }
      edge[s].clear();
      auto iter = nodes.find(s);
      if (iter != nodes.end()) nodes.erase(iter);
    };
  }
  return nodes.size();
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) nodes.insert(i);
  cnt = n;
  for (int i = 0; i < m; i++) {
    int s, t;
    cin >> s >> t;
    vE.push_back(Edge{s, t});
    edge[s].insert(t);
    edge[t].insert(s);
  }
  for (int i = 1; i <= n; i++)
    if (edge[i].size() < k) Q.push(i);
  int ret = do_remove();
  ans.push_back(ret);
  for (int i = vE.size() - 1; i >= 1; i--) {
    int s = vE[i].u;
    int t = vE[i].v;
    edge[s].erase(t);
    edge[t].erase(s);
    Q.push(s);
    Q.push(t);
    int ret = do_remove();
    ans.push_back(ret);
  }
  for (int i = ans.size() - 1; i >= 0; i--) {
    cout << ans[i] << endl;
  }
  return 0;
}

```