#include <bits/stdc++.h>
int n, m, k, u[200010], v[200010], t[200010];
std::set<int> next[200010], p;
inline void remove(int x) {
  if ((int)next[x].size() < k && p.erase(x)) {
    for (auto y : next[x]) {
      next[y].erase(x);
      remove(y);
    }
    next[x].clear();
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", u + i, v + i);
    next[u[i]].insert(v[i]);
    next[v[i]].insert(u[i]);
  }
  for (int i = 1; i <= n; i++) p.insert(i);
  for (int i = 1; i <= n; i++) remove(i);
  for (int i = m; i; i--) {
    t[i] = p.size();
    if (p.count(v[i]) && p.count(u[i])) {
      next[v[i]].erase(u[i]);
      next[u[i]].erase(v[i]);
      remove(u[i]);
      remove(v[i]);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", t[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>
int n, m, k, u[200010], v[200010], t[200010];
std::set<int> next[200010], p;
inline void remove(int x) {
  if ((int)next[x].size() < k && p.erase(x)) {
    for (auto y : next[x]) {
      next[y].erase(x);
      remove(y);
    }
    next[x].clear();
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", u + i, v + i);
    next[u[i]].insert(v[i]);
    next[v[i]].insert(u[i]);
  }
  for (int i = 1; i <= n; i++) p.insert(i);
  for (int i = 1; i <= n; i++) remove(i);
  for (int i = m; i; i--) {
    t[i] = p.size();
    if (p.count(v[i]) && p.count(u[i])) {
      next[v[i]].erase(u[i]);
      next[u[i]].erase(v[i]);
      remove(u[i]);
      remove(v[i]);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", t[i]);
  return 0;
}

```