#include <bits/stdc++.h>
using namespace std;
int n, m, fs;
std::vector<int> yol[200005];
pair<int, int> o[200005];
int ans, yaz[200005], say[200005], del[200005];
map<pair<int, int>, int> mp;
void sil(int k) {
  if (del[k] == 1) return;
  del[k] = 1;
  ans--;
  for (int i = 0; i < yol[k].size(); ++i) {
    if (mp[make_pair(k, yol[k][i])] || mp[make_pair(yol[k][i], k)]) continue;
    say[yol[k][i]]--;
    if (say[yol[k][i]] < fs && del[yol[k][i]] == 0) sil(yol[k][i]);
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &fs);
  ans = n;
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &o[i].first, &o[i].second);
    yol[o[i].first].push_back(o[i].second);
    yol[o[i].second].push_back(o[i].first);
    say[o[i].first]++;
    say[o[i].second]++;
  }
  for (int i = 1; i <= n; ++i) {
    if (say[i] < fs) {
      sil(i);
    }
  }
  for (int i = m - 1; i >= 0; --i) {
    yaz[i] = ans;
    if (del[o[i].first] == 0 && del[o[i].second] == 0) {
      say[o[i].first]--;
      say[o[i].second]--;
      mp[o[i]] = 1;
    }
    if (say[o[i].first] < fs) sil(o[i].first);
    if (say[o[i].second] < fs) sil(o[i].second);
  }
  for (int i = 0; i < m; ++i) printf("%d\n", yaz[i]);
}

### Prompt

Please formulate a CPP solution to 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;
int n, m, fs;
std::vector<int> yol[200005];
pair<int, int> o[200005];
int ans, yaz[200005], say[200005], del[200005];
map<pair<int, int>, int> mp;
void sil(int k) {
  if (del[k] == 1) return;
  del[k] = 1;
  ans--;
  for (int i = 0; i < yol[k].size(); ++i) {
    if (mp[make_pair(k, yol[k][i])] || mp[make_pair(yol[k][i], k)]) continue;
    say[yol[k][i]]--;
    if (say[yol[k][i]] < fs && del[yol[k][i]] == 0) sil(yol[k][i]);
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &fs);
  ans = n;
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &o[i].first, &o[i].second);
    yol[o[i].first].push_back(o[i].second);
    yol[o[i].second].push_back(o[i].first);
    say[o[i].first]++;
    say[o[i].second]++;
  }
  for (int i = 1; i <= n; ++i) {
    if (say[i] < fs) {
      sil(i);
    }
  }
  for (int i = m - 1; i >= 0; --i) {
    yaz[i] = ans;
    if (del[o[i].first] == 0 && del[o[i].second] == 0) {
      say[o[i].first]--;
      say[o[i].second]--;
      mp[o[i]] = 1;
    }
    if (say[o[i].first] < fs) sil(o[i].first);
    if (say[o[i].second] < fs) sil(o[i].second);
  }
  for (int i = 0; i < m; ++i) printf("%d\n", yaz[i]);
}

```