#include <bits/stdc++.h>
using namespace std;
int x[200010], y[200010];
set<int> s[200010];
queue<int> q, p;
bool inq[200010];
bool w[200010];
int ans[200010];
int n, m, k, sum;
void reset() {
  while (!q.empty()) q.pop();
  while (!p.empty()) inq[p.front()] = false, p.pop();
}
void BFS() {
  while (!q.empty()) {
    int now = q.front();
    w[now] = true;
    p.push(now);
    sum--;
    q.pop();
    for (set<int>::iterator i = s[now].begin(); i != s[now].end(); i++) {
      s[*i].erase(now);
      if (s[*i].size() < k && !inq[*i] && !w[*i]) inq[*i] = true, q.push(*i);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  sum = n;
  for (int i = 1; i <= m; i++)
    cin >> x[i] >> y[i], s[x[i]].insert(y[i]), s[y[i]].insert(x[i]);
  for (int i = 1; i <= n; i++)
    if (s[i].size() < k) inq[i] = true, q.push(i);
  BFS();
  for (int i = m; i > 0; i--) {
    ans[i] = sum;
    reset();
    s[x[i]].erase(y[i]);
    s[y[i]].erase(x[i]);
    if (s[x[i]].size() < k && !w[x[i]]) inq[x[i]] = true, q.push(x[i]);
    if (s[y[i]].size() < k && !w[y[i]]) inq[y[i]] = true, q.push(y[i]);
    BFS();
  }
  for (int i = 1; i <= m; 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 x[200010], y[200010];
set<int> s[200010];
queue<int> q, p;
bool inq[200010];
bool w[200010];
int ans[200010];
int n, m, k, sum;
void reset() {
  while (!q.empty()) q.pop();
  while (!p.empty()) inq[p.front()] = false, p.pop();
}
void BFS() {
  while (!q.empty()) {
    int now = q.front();
    w[now] = true;
    p.push(now);
    sum--;
    q.pop();
    for (set<int>::iterator i = s[now].begin(); i != s[now].end(); i++) {
      s[*i].erase(now);
      if (s[*i].size() < k && !inq[*i] && !w[*i]) inq[*i] = true, q.push(*i);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  sum = n;
  for (int i = 1; i <= m; i++)
    cin >> x[i] >> y[i], s[x[i]].insert(y[i]), s[y[i]].insert(x[i]);
  for (int i = 1; i <= n; i++)
    if (s[i].size() < k) inq[i] = true, q.push(i);
  BFS();
  for (int i = m; i > 0; i--) {
    ans[i] = sum;
    reset();
    s[x[i]].erase(y[i]);
    s[y[i]].erase(x[i]);
    if (s[x[i]].size() < k && !w[x[i]]) inq[x[i]] = true, q.push(x[i]);
    if (s[y[i]].size() < k && !w[y[i]]) inq[y[i]] = true, q.push(y[i]);
    BFS();
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  return 0;
}

```