#include <bits/stdc++.h>
using namespace std;
const int N = int(200000) + 5;
int t, n, m, q, k, cnt, sum, x, y, l, r;
int a[N], b[N], ans[N];
pair<int, int> edge[N];
vector<pair<int, int> > v[N];
set<pair<int, int> > s;
set<pair<int, int> >::iterator it;
void update(int I) {
  while (!s.empty() && (s.begin()->first) < k) {
    it = s.begin();
    x = it->second;
    for (int i = 0; i < v[x].size(); i++) {
      y = v[x][i].first;
      if (v[x][i].second >= I) continue;
      if (b[y]) {
        s.erase({a[y], y});
        a[y]--;
        s.insert({a[y], y});
      }
    }
    b[x] = 0;
    s.erase(it);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    edge[i].first = x;
    edge[i].second = y;
    a[x]++;
    a[y]++;
    v[x].push_back({y, i});
    v[y].push_back({x, i});
  }
  for (int i = 1; i <= n; i++) {
    s.insert({a[i], i});
    b[i] = 1;
  }
  update(1e6);
  for (int i = m; i >= 1; i--) {
    ans[i] = s.size();
    l = edge[i].first;
    r = edge[i].second;
    if (b[l] && b[r]) {
      s.erase({a[l], l});
      a[l]--;
      s.insert({a[l], l});
      s.erase({a[r], r});
      a[r]--;
      s.insert({a[r], r});
      update(i);
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
  return 0;
}

### Prompt

Generate a cpp solution 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;
const int N = int(200000) + 5;
int t, n, m, q, k, cnt, sum, x, y, l, r;
int a[N], b[N], ans[N];
pair<int, int> edge[N];
vector<pair<int, int> > v[N];
set<pair<int, int> > s;
set<pair<int, int> >::iterator it;
void update(int I) {
  while (!s.empty() && (s.begin()->first) < k) {
    it = s.begin();
    x = it->second;
    for (int i = 0; i < v[x].size(); i++) {
      y = v[x][i].first;
      if (v[x][i].second >= I) continue;
      if (b[y]) {
        s.erase({a[y], y});
        a[y]--;
        s.insert({a[y], y});
      }
    }
    b[x] = 0;
    s.erase(it);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    edge[i].first = x;
    edge[i].second = y;
    a[x]++;
    a[y]++;
    v[x].push_back({y, i});
    v[y].push_back({x, i});
  }
  for (int i = 1; i <= n; i++) {
    s.insert({a[i], i});
    b[i] = 1;
  }
  update(1e6);
  for (int i = m; i >= 1; i--) {
    ans[i] = s.size();
    l = edge[i].first;
    r = edge[i].second;
    if (b[l] && b[r]) {
      s.erase({a[l], l});
      a[l]--;
      s.insert({a[l], l});
      s.erase({a[r], r});
      a[r]--;
      s.insert({a[r], r});
      update(i);
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
  return 0;
}

```