#include <bits/stdc++.h>
using namespace std;
int main(int agrc, char* argv[]) {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  std::vector<std::vector<int>> edges(n + 1);
  int selected[n];
  for (int i = 0; i < n; ++i) {
    selected[i] = 1;
  }
  int total = n;
  int friends[m][2];
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &friends[i][0], &friends[i][1]);
    --friends[i][0];
    --friends[i][1];
    edges[friends[i][0]].push_back(friends[i][1]);
    edges[friends[i][1]].push_back(friends[i][0]);
  }
  int friends_going[n];
  for (int i = 0; i < n; ++i) {
    friends_going[i] = edges[i].size();
  }
  for (int i = 0; i < n; ++i) {
    std::stack<int> rejected;
    rejected.push(i);
    while (!rejected.empty()) {
      int index = rejected.top();
      rejected.pop();
      if (!(selected[index] == 1 && friends_going[index] < k)) {
        continue;
      }
      selected[index] = 0;
      --total;
      for (int j : edges[index]) {
        --friends_going[j];
        rejected.push(j);
      }
    }
  }
  int answer[m];
  answer[m - 1] = total;
  for (int i = m - 2; i >= 0; --i) {
    int first = friends[i + 1][0], second = friends[i + 1][1];
    edges[first].pop_back();
    edges[second].pop_back();
    if (selected[second] == 1) {
      --friends_going[first];
    }
    if (selected[first]) {
      --friends_going[second];
    }
    std::stack<int> rejected;
    rejected.push(first);
    rejected.push(second);
    while (!rejected.empty()) {
      int index = rejected.top();
      rejected.pop();
      if (!(selected[index] == 1 && friends_going[index] < k)) {
        continue;
      }
      selected[index] = 0;
      --total;
      for (int j : edges[index]) {
        --friends_going[j];
        rejected.push(j);
      }
    }
    answer[i] = total;
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", answer[i]);
  }
}

### 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>
using namespace std;
int main(int agrc, char* argv[]) {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  std::vector<std::vector<int>> edges(n + 1);
  int selected[n];
  for (int i = 0; i < n; ++i) {
    selected[i] = 1;
  }
  int total = n;
  int friends[m][2];
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &friends[i][0], &friends[i][1]);
    --friends[i][0];
    --friends[i][1];
    edges[friends[i][0]].push_back(friends[i][1]);
    edges[friends[i][1]].push_back(friends[i][0]);
  }
  int friends_going[n];
  for (int i = 0; i < n; ++i) {
    friends_going[i] = edges[i].size();
  }
  for (int i = 0; i < n; ++i) {
    std::stack<int> rejected;
    rejected.push(i);
    while (!rejected.empty()) {
      int index = rejected.top();
      rejected.pop();
      if (!(selected[index] == 1 && friends_going[index] < k)) {
        continue;
      }
      selected[index] = 0;
      --total;
      for (int j : edges[index]) {
        --friends_going[j];
        rejected.push(j);
      }
    }
  }
  int answer[m];
  answer[m - 1] = total;
  for (int i = m - 2; i >= 0; --i) {
    int first = friends[i + 1][0], second = friends[i + 1][1];
    edges[first].pop_back();
    edges[second].pop_back();
    if (selected[second] == 1) {
      --friends_going[first];
    }
    if (selected[first]) {
      --friends_going[second];
    }
    std::stack<int> rejected;
    rejected.push(first);
    rejected.push(second);
    while (!rejected.empty()) {
      int index = rejected.top();
      rejected.pop();
      if (!(selected[index] == 1 && friends_going[index] < k)) {
        continue;
      }
      selected[index] = 0;
      --total;
      for (int j : edges[index]) {
        --friends_going[j];
        rejected.push(j);
      }
    }
    answer[i] = total;
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", answer[i]);
  }
}

```