#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("-ffloat-store")
using namespace std;
const int MN = 200010;
set<int> e[MN];
int from[MN], to[MN], deg[MN], vis[MN], res[MN];
int n, m, k, cnt;
void deletev(int v) {
  if (deg[v] >= k || vis[v]) return;
  queue<int> q;
  q.push(v);
  vis[v] = 1;
  --cnt;
  while (!q.empty()) {
    int top = q.front();
    q.pop();
    for (auto& c : e[top]) {
      --deg[c];
      if (deg[c] < k && !vis[c]) {
        q.push(c);
        vis[c] = 1;
        --cnt;
      }
    }
  }
}
struct E {
  void solve(std::istream& cin, std::ostream& cout) {
    cin >> n >> m >> k;
    memset(deg, (0), sizeof(deg));
    memset(from, (0), sizeof(from));
    memset(to, (0), sizeof(to));
    memset(deg, (0), sizeof(deg));
    memset(vis, (0), sizeof(vis));
    memset(res, (0), sizeof(res));
    e->clear();
    for (int i = (0); i < (m); ++i) {
      cin >> from[i] >> to[i];
      --from[i], --to[i];
      ++deg[from[i]], ++deg[to[i]];
      e[from[i]].insert(to[i]);
      e[to[i]].insert(from[i]);
    }
    cnt = n;
    for (int i = (0); i < (n); ++i) {
      deletev(i);
    }
    res[m] = cnt;
    for (int i = (m - 1); i >= (0); --i) {
      if (!vis[from[i]]) --deg[to[i]];
      if (!vis[to[i]]) --deg[from[i]];
      e[from[i]].erase(to[i]);
      e[to[i]].erase(from[i]);
      deletev(from[i]);
      deletev(to[i]);
      res[i] = cnt;
    }
    for (int i = (1); i <= (m); ++i) cout << res[i] << "\n";
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  E solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

### Prompt

Develop a solution in cpp 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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("-ffloat-store")
using namespace std;
const int MN = 200010;
set<int> e[MN];
int from[MN], to[MN], deg[MN], vis[MN], res[MN];
int n, m, k, cnt;
void deletev(int v) {
  if (deg[v] >= k || vis[v]) return;
  queue<int> q;
  q.push(v);
  vis[v] = 1;
  --cnt;
  while (!q.empty()) {
    int top = q.front();
    q.pop();
    for (auto& c : e[top]) {
      --deg[c];
      if (deg[c] < k && !vis[c]) {
        q.push(c);
        vis[c] = 1;
        --cnt;
      }
    }
  }
}
struct E {
  void solve(std::istream& cin, std::ostream& cout) {
    cin >> n >> m >> k;
    memset(deg, (0), sizeof(deg));
    memset(from, (0), sizeof(from));
    memset(to, (0), sizeof(to));
    memset(deg, (0), sizeof(deg));
    memset(vis, (0), sizeof(vis));
    memset(res, (0), sizeof(res));
    e->clear();
    for (int i = (0); i < (m); ++i) {
      cin >> from[i] >> to[i];
      --from[i], --to[i];
      ++deg[from[i]], ++deg[to[i]];
      e[from[i]].insert(to[i]);
      e[to[i]].insert(from[i]);
    }
    cnt = n;
    for (int i = (0); i < (n); ++i) {
      deletev(i);
    }
    res[m] = cnt;
    for (int i = (m - 1); i >= (0); --i) {
      if (!vis[from[i]]) --deg[to[i]];
      if (!vis[to[i]]) --deg[from[i]];
      e[from[i]].erase(to[i]);
      e[to[i]].erase(from[i]);
      deletev(from[i]);
      deletev(to[i]);
      res[i] = cnt;
    }
    for (int i = (1); i <= (m); ++i) cout << res[i] << "\n";
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  E solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

```