#include <bits/stdc++.h>
using namespace std;
pair<int, int> v[200100];
int lv[200100];
vector<int> gr[200100];
int in[200100];
vector<int> go;
map<pair<int, int>, int> taiat;
int n, m, k, cont;
void dfs(int nod) {
  in[nod] = 0;
  cont--;
  for (auto &x : gr[nod]) {
    if (taiat[{nod, x}]) {
      continue;
    }
    taiat[{nod, x}] = 1;
    taiat[{x, nod}] = 1;
    lv[x]--;
    if (in[x] && lv[x] < k) {
      dfs(x);
    }
  }
}
vector<int> ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  cont = n;
  for (int i = 1; i <= m; i++) {
    cin >> v[i].first >> v[i].second;
    gr[v[i].first].push_back(v[i].second);
    gr[v[i].second].push_back(v[i].first);
    lv[v[i].first]++;
    lv[v[i].second]++;
  }
  for (int i = 1; i <= n; i++) {
    in[i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (lv[i] < k) {
      go.push_back(i);
    }
  }
  for (auto &x : go) {
    if (in[x]) {
      dfs(x);
    }
  }
  for (int i = m; i >= 1; i--) {
    ans.push_back(cont);
    if (taiat[{v[i].first, v[i].second}]) {
      continue;
    }
    taiat[{v[i].first, v[i].second}] = 1;
    taiat[{v[i].second, v[i].first}] = 1;
    lv[v[i].first]--;
    lv[v[i].second]--;
    if (lv[v[i].first] < k && in[v[i].first]) {
      dfs(v[i].first);
    }
    if (lv[v[i].second] < k && in[v[i].second]) {
      dfs(v[i].second);
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto &x : ans) {
    cout << x << '\n';
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
pair<int, int> v[200100];
int lv[200100];
vector<int> gr[200100];
int in[200100];
vector<int> go;
map<pair<int, int>, int> taiat;
int n, m, k, cont;
void dfs(int nod) {
  in[nod] = 0;
  cont--;
  for (auto &x : gr[nod]) {
    if (taiat[{nod, x}]) {
      continue;
    }
    taiat[{nod, x}] = 1;
    taiat[{x, nod}] = 1;
    lv[x]--;
    if (in[x] && lv[x] < k) {
      dfs(x);
    }
  }
}
vector<int> ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  cont = n;
  for (int i = 1; i <= m; i++) {
    cin >> v[i].first >> v[i].second;
    gr[v[i].first].push_back(v[i].second);
    gr[v[i].second].push_back(v[i].first);
    lv[v[i].first]++;
    lv[v[i].second]++;
  }
  for (int i = 1; i <= n; i++) {
    in[i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (lv[i] < k) {
      go.push_back(i);
    }
  }
  for (auto &x : go) {
    if (in[x]) {
      dfs(x);
    }
  }
  for (int i = m; i >= 1; i--) {
    ans.push_back(cont);
    if (taiat[{v[i].first, v[i].second}]) {
      continue;
    }
    taiat[{v[i].first, v[i].second}] = 1;
    taiat[{v[i].second, v[i].first}] = 1;
    lv[v[i].first]--;
    lv[v[i].second]--;
    if (lv[v[i].first] < k && in[v[i].first]) {
      dfs(v[i].first);
    }
    if (lv[v[i].second] < k && in[v[i].second]) {
      dfs(v[i].second);
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto &x : ans) {
    cout << x << '\n';
  }
  return 0;
}

```