#include <bits/stdc++.h>
const int ms = 200200;
int n, m, k;
int ans[ms];
std::pair<int, int> edges[ms];
std::vector<int> use[ms];
int deg[ms];
bool del[ms];
int main() {
  std::cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    use[u].push_back(i);
    use[v].push_back(i);
    deg[u]++;
    deg[v]++;
    edges[i] = std::pair<int, int>(u, v);
  }
  std::queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (deg[i] < k) {
      q.push(i);
    }
  }
  del[m] = true;
  int cur_ans = n;
  for (int i = m; i >= 0; i--) {
    if (!del[i]) {
      int u = edges[i].first, v = edges[i].second;
      if (deg[u] == k) {
        q.push(u);
      }
      if (deg[v] == k) {
        q.push(v);
      }
      deg[u]--;
      deg[v]--;
      del[i] = true;
    }
    while (!q.empty()) {
      cur_ans--;
      int on = q.front();
      q.pop();
      for (auto e : use[on]) {
        if (!del[e]) {
          int u = edges[e].first, v = edges[e].second;
          if (deg[u] == k) {
            q.push(u);
          }
          if (deg[v] == k) {
            q.push(v);
          }
          deg[u]--;
          deg[v]--;
          del[e] = true;
        }
      }
    }
    ans[i] = cur_ans;
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", ans[i + 1]);
  }
}

### 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>
const int ms = 200200;
int n, m, k;
int ans[ms];
std::pair<int, int> edges[ms];
std::vector<int> use[ms];
int deg[ms];
bool del[ms];
int main() {
  std::cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    use[u].push_back(i);
    use[v].push_back(i);
    deg[u]++;
    deg[v]++;
    edges[i] = std::pair<int, int>(u, v);
  }
  std::queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (deg[i] < k) {
      q.push(i);
    }
  }
  del[m] = true;
  int cur_ans = n;
  for (int i = m; i >= 0; i--) {
    if (!del[i]) {
      int u = edges[i].first, v = edges[i].second;
      if (deg[u] == k) {
        q.push(u);
      }
      if (deg[v] == k) {
        q.push(v);
      }
      deg[u]--;
      deg[v]--;
      del[i] = true;
    }
    while (!q.empty()) {
      cur_ans--;
      int on = q.front();
      q.pop();
      for (auto e : use[on]) {
        if (!del[e]) {
          int u = edges[e].first, v = edges[e].second;
          if (deg[u] == k) {
            q.push(u);
          }
          if (deg[v] == k) {
            q.push(v);
          }
          deg[u]--;
          deg[v]--;
          del[e] = true;
        }
      }
    }
    ans[i] = cur_ans;
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", ans[i + 1]);
  }
}

```