#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int n, m, k, X[MAXN], Y[MAXN], deg[MAXN], cnt;
vector<pair<int, int> > E[MAXN];
bool removed[MAXN];
void ukloni(int x, int tijme) {
  if (!removed[x] && deg[x] < k) {
    removed[x] = true;
    cnt--;
    for (auto e : E[x])
      if (e.second < tijme) {
        deg[e.first]--;
        ukloni(e.first, tijme);
      }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    cin >> X[i] >> Y[i];
    E[X[i]].push_back(pair<int, int>(Y[i], i));
    E[Y[i]].push_back(pair<int, int>(X[i], i));
    deg[X[i]]++;
    deg[Y[i]]++;
  }
  cnt = n;
  for (int i = 1; i < n + 1; ++i) ukloni(i, MAXN);
  vector<int> sol;
  for (int i = m - 1; i >= 0; --i) {
    sol.push_back(cnt);
    if (!removed[X[i]] && !removed[Y[i]]) {
      deg[X[i]]--;
      deg[Y[i]]--;
      ukloni(X[i], i);
      ukloni(Y[i], i);
    }
  }
  for (int i = ((int)sol.size()) - 1; i >= 0; --i) cout << sol[i] << "\n";
}

### Prompt

In cpp, your task is to solve 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;
const int MAXN = 2e5 + 5;
int n, m, k, X[MAXN], Y[MAXN], deg[MAXN], cnt;
vector<pair<int, int> > E[MAXN];
bool removed[MAXN];
void ukloni(int x, int tijme) {
  if (!removed[x] && deg[x] < k) {
    removed[x] = true;
    cnt--;
    for (auto e : E[x])
      if (e.second < tijme) {
        deg[e.first]--;
        ukloni(e.first, tijme);
      }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    cin >> X[i] >> Y[i];
    E[X[i]].push_back(pair<int, int>(Y[i], i));
    E[Y[i]].push_back(pair<int, int>(X[i], i));
    deg[X[i]]++;
    deg[Y[i]]++;
  }
  cnt = n;
  for (int i = 1; i < n + 1; ++i) ukloni(i, MAXN);
  vector<int> sol;
  for (int i = m - 1; i >= 0; --i) {
    sol.push_back(cnt);
    if (!removed[X[i]] && !removed[Y[i]]) {
      deg[X[i]]--;
      deg[Y[i]]--;
      ukloni(X[i], i);
      ukloni(Y[i], i);
    }
  }
  for (int i = ((int)sol.size()) - 1; i >= 0; --i) cout << sol[i] << "\n";
}

```