#include <bits/stdc++.h>
using namespace std;
int n, m, k;
set<pair<int, int>> S;
vector<int> A;
vector<pair<int, int>> R;
vector<vector<int>> G;
set<pair<int, int>> B;
int main() {
  cin >> n >> m >> k;
  R.assign(m, {0, 0});
  A.assign(n, 0);
  G.assign(n, vector<int>(0, 0));
  for (auto &i : R) {
    cin >> i.first >> i.second;
    i.first--;
    i.second--;
    G[i.first].push_back(i.second);
    G[i.second].push_back(i.first);
    A[i.first]++;
    A[i.second]++;
  }
  for (int i = 0; i < n; i++) {
    S.insert({A[i], i});
  }
  while (S.size() && (*S.begin()).first < k) {
    int x = (*S.begin()).second;
    S.erase(S.begin());
    for (int y : G[x]) {
      if (A[y] == -1) continue;
      S.erase(S.find({A[y], y}));
      A[y]--;
      S.insert({A[y], y});
    }
    A[x] = -1;
  }
  vector<int> Ans(m, 0);
  for (int i = m - 1; i >= 0; i--) {
    Ans[i] = S.size();
    int x, y;
    x = R[i].first;
    y = R[i].second;
    B.insert({x, y});
    B.insert({y, x});
    if (A[x] != -1 && A[y] != -1) {
      S.erase(S.find({A[x], x}));
      A[x]--;
      S.insert({A[x], x});
      S.erase(S.find({A[y], y}));
      A[y]--;
      S.insert({A[y], y});
    }
    while (!S.empty() && (*S.begin()).first < k) {
      int X = (*S.begin()).second;
      S.erase(S.begin());
      for (int Y : G[X]) {
        if (A[Y] == -1 || B.count({X, Y})) continue;
        S.erase(S.find({A[Y], Y}));
        A[Y]--;
        S.insert({A[Y], Y});
      }
      A[X] = -1;
    }
  }
  for (int i : Ans) {
    cout << 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>
using namespace std;
int n, m, k;
set<pair<int, int>> S;
vector<int> A;
vector<pair<int, int>> R;
vector<vector<int>> G;
set<pair<int, int>> B;
int main() {
  cin >> n >> m >> k;
  R.assign(m, {0, 0});
  A.assign(n, 0);
  G.assign(n, vector<int>(0, 0));
  for (auto &i : R) {
    cin >> i.first >> i.second;
    i.first--;
    i.second--;
    G[i.first].push_back(i.second);
    G[i.second].push_back(i.first);
    A[i.first]++;
    A[i.second]++;
  }
  for (int i = 0; i < n; i++) {
    S.insert({A[i], i});
  }
  while (S.size() && (*S.begin()).first < k) {
    int x = (*S.begin()).second;
    S.erase(S.begin());
    for (int y : G[x]) {
      if (A[y] == -1) continue;
      S.erase(S.find({A[y], y}));
      A[y]--;
      S.insert({A[y], y});
    }
    A[x] = -1;
  }
  vector<int> Ans(m, 0);
  for (int i = m - 1; i >= 0; i--) {
    Ans[i] = S.size();
    int x, y;
    x = R[i].first;
    y = R[i].second;
    B.insert({x, y});
    B.insert({y, x});
    if (A[x] != -1 && A[y] != -1) {
      S.erase(S.find({A[x], x}));
      A[x]--;
      S.insert({A[x], x});
      S.erase(S.find({A[y], y}));
      A[y]--;
      S.insert({A[y], y});
    }
    while (!S.empty() && (*S.begin()).first < k) {
      int X = (*S.begin()).second;
      S.erase(S.begin());
      for (int Y : G[X]) {
        if (A[Y] == -1 || B.count({X, Y})) continue;
        S.erase(S.find({A[Y], Y}));
        A[Y]--;
        S.insert({A[Y], Y});
      }
      A[X] = -1;
    }
  }
  for (int i : Ans) {
    cout << i << "\n";
  }
}

```