#include <bits/stdc++.h>
using namespace std;
const int mx = 200 * 1000 + 5;
vector<set<int> > vec(mx);
int arr1[mx], arr2[mx];
int main() {
  int i, j, k, n, m, x, y, p, q, l;
  scanf("%d %d %d", &n, &m, &k);
  int ans = 0;
  for (i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    arr1[i] = x;
    arr2[i] = y;
    vec[x].insert(y);
    vec[y].insert(x);
  }
  ans = n;
  queue<int> ajairra;
  bool vis[n + 5];
  memset(vis, 0, sizeof(vis));
  for (i = 1; i <= n; i++) {
    p = vec[i].size();
    if (p < k) {
      vis[i] = 1;
      ans--;
      ajairra.push(i);
    }
  }
  vector<int> vv;
  set<int>::iterator itr;
  for (i = m - 1; i >= 0; i--) {
    while (!ajairra.empty()) {
      p = ajairra.front();
      ajairra.pop();
      for (itr = vec[p].begin(); itr != vec[p].end(); itr++) {
        q = *itr;
        vec[q].erase(p);
        l = vec[q].size();
        if (l < k && !vis[q]) {
          vis[q] = 1;
          ans--;
          ajairra.push(q);
        }
      }
      vec[p].clear();
    }
    vv.push_back(ans);
    p = arr1[i];
    q = arr2[i];
    vec[p].erase(q);
    vec[q].erase(p);
    l = vec[p].size();
    if (l < k && !vis[p]) {
      vis[p] = 1;
      ans--;
      ajairra.push(p);
    }
    l = vec[q].size();
    if (l < k && !vis[q]) {
      vis[q] = 1;
      ans--;
      ajairra.push(q);
    }
  }
  for (i = m - 1; i >= 0; i--) {
    printf("%d\n", vv[i]);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int mx = 200 * 1000 + 5;
vector<set<int> > vec(mx);
int arr1[mx], arr2[mx];
int main() {
  int i, j, k, n, m, x, y, p, q, l;
  scanf("%d %d %d", &n, &m, &k);
  int ans = 0;
  for (i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    arr1[i] = x;
    arr2[i] = y;
    vec[x].insert(y);
    vec[y].insert(x);
  }
  ans = n;
  queue<int> ajairra;
  bool vis[n + 5];
  memset(vis, 0, sizeof(vis));
  for (i = 1; i <= n; i++) {
    p = vec[i].size();
    if (p < k) {
      vis[i] = 1;
      ans--;
      ajairra.push(i);
    }
  }
  vector<int> vv;
  set<int>::iterator itr;
  for (i = m - 1; i >= 0; i--) {
    while (!ajairra.empty()) {
      p = ajairra.front();
      ajairra.pop();
      for (itr = vec[p].begin(); itr != vec[p].end(); itr++) {
        q = *itr;
        vec[q].erase(p);
        l = vec[q].size();
        if (l < k && !vis[q]) {
          vis[q] = 1;
          ans--;
          ajairra.push(q);
        }
      }
      vec[p].clear();
    }
    vv.push_back(ans);
    p = arr1[i];
    q = arr2[i];
    vec[p].erase(q);
    vec[q].erase(p);
    l = vec[p].size();
    if (l < k && !vis[p]) {
      vis[p] = 1;
      ans--;
      ajairra.push(p);
    }
    l = vec[q].size();
    if (l < k && !vis[q]) {
      vis[q] = 1;
      ans--;
      ajairra.push(q);
    }
  }
  for (i = m - 1; i >= 0; i--) {
    printf("%d\n", vv[i]);
  }
  return 0;
}

```