#include <bits/stdc++.h>
const int N = 2e5 + 5;
using namespace std;
int n, m, k;
int res[N];
int x[N], y[N];
int bac[N];
set<int> a[N];
set<pair<int, int> > S;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> x[i] >> y[i];
    a[x[i]].insert(y[i]);
    a[y[i]].insert(x[i]);
    bac[x[i]]++;
    bac[y[i]]++;
  }
  for (int i = 1; i <= n; i++)
    if (bac[i] > 0) S.insert(make_pair(bac[i], i));
  for (int i = m; i >= 1; i--) {
    while (S.size() && S.begin()->first < k) {
      int cc = S.begin()->first;
      int u = S.begin()->second;
      for (int v : a[u]) {
        S.erase(make_pair(bac[v], v));
        bac[v]--;
        if (bac[v] > 0) S.insert(make_pair(bac[v], v));
        a[v].erase(u);
      }
      a[u].clear();
      bac[u] = 0;
      S.erase(make_pair(cc, u));
    }
    res[i] = S.size();
    if (a[y[i]].find(x[i]) != a[y[i]].end()) {
      S.erase(make_pair(bac[x[i]], x[i]));
      bac[x[i]]--;
      if (bac[x[i]] > 0) S.insert(make_pair(bac[x[i]], x[i]));
      S.erase(make_pair(bac[y[i]], y[i]));
      bac[y[i]]--;
      if (bac[y[i]] > 0) S.insert(make_pair(bac[y[i]], y[i]));
      a[x[i]].erase(y[i]);
      a[y[i]].erase(x[i]);
    }
  }
  for (int i = 1; i <= m; i++) cout << res[i] << '\n';
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
const int N = 2e5 + 5;
using namespace std;
int n, m, k;
int res[N];
int x[N], y[N];
int bac[N];
set<int> a[N];
set<pair<int, int> > S;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> x[i] >> y[i];
    a[x[i]].insert(y[i]);
    a[y[i]].insert(x[i]);
    bac[x[i]]++;
    bac[y[i]]++;
  }
  for (int i = 1; i <= n; i++)
    if (bac[i] > 0) S.insert(make_pair(bac[i], i));
  for (int i = m; i >= 1; i--) {
    while (S.size() && S.begin()->first < k) {
      int cc = S.begin()->first;
      int u = S.begin()->second;
      for (int v : a[u]) {
        S.erase(make_pair(bac[v], v));
        bac[v]--;
        if (bac[v] > 0) S.insert(make_pair(bac[v], v));
        a[v].erase(u);
      }
      a[u].clear();
      bac[u] = 0;
      S.erase(make_pair(cc, u));
    }
    res[i] = S.size();
    if (a[y[i]].find(x[i]) != a[y[i]].end()) {
      S.erase(make_pair(bac[x[i]], x[i]));
      bac[x[i]]--;
      if (bac[x[i]] > 0) S.insert(make_pair(bac[x[i]], x[i]));
      S.erase(make_pair(bac[y[i]], y[i]));
      bac[y[i]]--;
      if (bac[y[i]] > 0) S.insert(make_pair(bac[y[i]], y[i]));
      a[x[i]].erase(y[i]);
      a[y[i]].erase(x[i]);
    }
  }
  for (int i = 1; i <= m; i++) cout << res[i] << '\n';
}

```