#include <bits/stdc++.h>
using namespace std;
int n, m, k, a, b, x;
vector<int> edge[200005];
pair<int, int> amie[200005];
int deg[200005], ans[200005];
set<pair<int, int> > s, r;
void go(int y, int z) {
  while (!s.empty() && s.begin()->first < k) {
    a = s.begin()->second;
    b = (int)edge[a].size();
    s.erase(s.begin());
    for (int i = 0; i < b; ++i) {
      if ((a == y && edge[a][i] == z) || (a == z && edge[a][i] == y)) continue;
      if (s.find(pair<int, int>(deg[edge[a][i]], edge[a][i])) != s.end() &&
          r.find(pair<int, int>(a, edge[a][i])) == r.end() &&
          r.find(pair<int, int>(edge[a][i], a)) == r.end()) {
        s.erase(s.find(pair<int, int>(deg[edge[a][i]], edge[a][i])));
        deg[edge[a][i]]--;
        s.insert(pair<int, int>(deg[edge[a][i]], edge[a][i]));
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &amie[i].first, &amie[i].second);
    edge[amie[i].first].push_back(amie[i].second);
    edge[amie[i].second].push_back(amie[i].first);
    deg[amie[i].first]++;
    deg[amie[i].second]++;
  }
  for (int i = 1; i <= n; ++i) s.insert(pair<int, int>(deg[i], i));
  go(-1, -1);
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = (int)s.size();
    if (s.find(pair<int, int>(deg[amie[i].first], amie[i].first)) != s.end() &&
        s.find(pair<int, int>(deg[amie[i].second], amie[i].second)) !=
            s.end()) {
      s.erase(s.find(pair<int, int>(deg[amie[i].first], amie[i].first)));
      s.erase(s.find(pair<int, int>(deg[amie[i].second], amie[i].second)));
      deg[amie[i].first]--;
      deg[amie[i].second]--;
      s.insert(pair<int, int>(deg[amie[i].first], amie[i].first));
      s.insert(pair<int, int>(deg[amie[i].second], amie[i].second));
      r.insert(amie[i]);
      go(amie[i].first, amie[i].second);
    }
  }
  for (int i = 0; i < m; ++i) {
    printf("%d\n", ans[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;
int n, m, k, a, b, x;
vector<int> edge[200005];
pair<int, int> amie[200005];
int deg[200005], ans[200005];
set<pair<int, int> > s, r;
void go(int y, int z) {
  while (!s.empty() && s.begin()->first < k) {
    a = s.begin()->second;
    b = (int)edge[a].size();
    s.erase(s.begin());
    for (int i = 0; i < b; ++i) {
      if ((a == y && edge[a][i] == z) || (a == z && edge[a][i] == y)) continue;
      if (s.find(pair<int, int>(deg[edge[a][i]], edge[a][i])) != s.end() &&
          r.find(pair<int, int>(a, edge[a][i])) == r.end() &&
          r.find(pair<int, int>(edge[a][i], a)) == r.end()) {
        s.erase(s.find(pair<int, int>(deg[edge[a][i]], edge[a][i])));
        deg[edge[a][i]]--;
        s.insert(pair<int, int>(deg[edge[a][i]], edge[a][i]));
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &amie[i].first, &amie[i].second);
    edge[amie[i].first].push_back(amie[i].second);
    edge[amie[i].second].push_back(amie[i].first);
    deg[amie[i].first]++;
    deg[amie[i].second]++;
  }
  for (int i = 1; i <= n; ++i) s.insert(pair<int, int>(deg[i], i));
  go(-1, -1);
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = (int)s.size();
    if (s.find(pair<int, int>(deg[amie[i].first], amie[i].first)) != s.end() &&
        s.find(pair<int, int>(deg[amie[i].second], amie[i].second)) !=
            s.end()) {
      s.erase(s.find(pair<int, int>(deg[amie[i].first], amie[i].first)));
      s.erase(s.find(pair<int, int>(deg[amie[i].second], amie[i].second)));
      deg[amie[i].first]--;
      deg[amie[i].second]--;
      s.insert(pair<int, int>(deg[amie[i].first], amie[i].first));
      s.insert(pair<int, int>(deg[amie[i].second], amie[i].second));
      r.insert(amie[i]);
      go(amie[i].first, amie[i].second);
    }
  }
  for (int i = 0; i < m; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```