#include <bits/stdc++.h>
using namespace std;
int n, m, k, x, y, deg[200010];
set<pair<int, int>> s;
pair<int, int> p[200010];
vector<int> rez;
set<int> v[200010];
map<bool, int> mp[200010];
void rem() {
  while (!s.empty() and s.begin()->first < k) {
    x = s.begin()->second;
    for (auto it : v[x]) {
      s.erase({deg[it], it});
      --deg[it];
      v[it].erase(x);
      if (deg[it] > 0) s.insert({deg[it], it});
    }
    s.erase({deg[x], x});
    deg[x] = 0;
    v[x].clear();
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> p[i].first >> p[i].second;
    deg[p[i].first]++, deg[p[i].second]++;
    v[p[i].first].insert(p[i].second), v[p[i].second].insert(p[i].first);
  }
  for (int i = 1; i <= n; i++) s.insert({deg[i], i});
  for (int i = m - 1; i >= 0; i--) {
    rem();
    rez.push_back((int)s.size());
    x = p[i].first, y = p[i].second;
    if (deg[x] == 0 or deg[y] == 0) continue;
    v[x].erase(y), v[y].erase(x);
    s.erase({deg[x], x}), s.erase({deg[y], y});
    --deg[x], --deg[y];
    if (deg[x] > 0) s.insert({deg[x], x});
    if (deg[y] > 0) s.insert({deg[y], y});
  }
  reverse(rez.begin(), rez.end());
  for (auto ans : rez) cout << ans << endl;
  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;
int n, m, k, x, y, deg[200010];
set<pair<int, int>> s;
pair<int, int> p[200010];
vector<int> rez;
set<int> v[200010];
map<bool, int> mp[200010];
void rem() {
  while (!s.empty() and s.begin()->first < k) {
    x = s.begin()->second;
    for (auto it : v[x]) {
      s.erase({deg[it], it});
      --deg[it];
      v[it].erase(x);
      if (deg[it] > 0) s.insert({deg[it], it});
    }
    s.erase({deg[x], x});
    deg[x] = 0;
    v[x].clear();
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> p[i].first >> p[i].second;
    deg[p[i].first]++, deg[p[i].second]++;
    v[p[i].first].insert(p[i].second), v[p[i].second].insert(p[i].first);
  }
  for (int i = 1; i <= n; i++) s.insert({deg[i], i});
  for (int i = m - 1; i >= 0; i--) {
    rem();
    rez.push_back((int)s.size());
    x = p[i].first, y = p[i].second;
    if (deg[x] == 0 or deg[y] == 0) continue;
    v[x].erase(y), v[y].erase(x);
    s.erase({deg[x], x}), s.erase({deg[y], y});
    --deg[x], --deg[y];
    if (deg[x] > 0) s.insert({deg[x], x});
    if (deg[y] > 0) s.insert({deg[y], y});
  }
  reverse(rez.begin(), rez.end());
  for (auto ans : rez) cout << ans << endl;
  return 0;
}

```