#include <bits/stdc++.h>
using namespace std;
set<int> tt[200005];
set<int> ans;
vector<pair<int, int>> ve;
int n, m, k, x, y;
int temp[200005];
void del(int x) {
  if (tt[x].size() < k && ans.find(x) != ans.end()) {
    ans.erase(x);
    for (auto i : tt[x]) {
      tt[i].erase(x);
      del(i);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    tt[x].insert(y);
    tt[y].insert(x);
    ve.push_back(make_pair(x, y));
  }
  for (int i = 1; i <= n; i++) {
    ans.insert(i);
  }
  for (int i = 1; i <= n; i++) {
    del(i);
  }
  for (int i = m - 1; i >= 0; i--) {
    temp[i] = ans.size();
    tt[ve[i].first].erase(ve[i].second);
    tt[ve[i].second].erase(ve[i].first);
    del(ve[i].first);
    del(ve[i].second);
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", temp[i]);
  }
  return 0;
}

### 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;
set<int> tt[200005];
set<int> ans;
vector<pair<int, int>> ve;
int n, m, k, x, y;
int temp[200005];
void del(int x) {
  if (tt[x].size() < k && ans.find(x) != ans.end()) {
    ans.erase(x);
    for (auto i : tt[x]) {
      tt[i].erase(x);
      del(i);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    tt[x].insert(y);
    tt[y].insert(x);
    ve.push_back(make_pair(x, y));
  }
  for (int i = 1; i <= n; i++) {
    ans.insert(i);
  }
  for (int i = 1; i <= n; i++) {
    del(i);
  }
  for (int i = m - 1; i >= 0; i--) {
    temp[i] = ans.size();
    tt[ve[i].first].erase(ve[i].second);
    tt[ve[i].second].erase(ve[i].first);
    del(ve[i].first);
    del(ve[i].second);
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", temp[i]);
  }
  return 0;
}

```