#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > adj;
set<long long> st;
vector<long long> deg;
long long k;
set<pair<long long, long long> > edges;
void remove(long long r) {
  st.erase(r);
  for (long long i = 0; i < adj[r].size(); i++) {
    long long c = adj[r][i];
    if (st.find(c) == st.end()) continue;
    if (edges.find(make_pair(r, c)) == edges.end()) continue;
    edges.erase(make_pair(r, c));
    edges.erase(make_pair(c, r));
    deg[c]--;
    if (deg[c] < k) remove(c);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m >> k;
  vector<long long> x(m);
  vector<long long> y(m);
  adj.resize(n);
  deg.resize(n, 0);
  for (long long i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    adj[x[i]].push_back(y[i]);
    adj[y[i]].push_back(x[i]);
    edges.insert(make_pair(x[i], y[i]));
    edges.insert(make_pair(y[i], x[i]));
    deg[x[i]]++;
    deg[y[i]]++;
  }
  for (long long i = 0; i < n; i++) {
    st.insert(i);
  }
  for (long long i = 0; i < n; i++) {
    if (st.find(i) == st.end()) continue;
    if (deg[i] < k) remove(i);
  }
  vector<long long> res;
  res.push_back(st.size());
  for (long long i = m - 1; i >= 1; i--) {
    if (st.find(x[i]) == st.end() || st.find(y[i]) == st.end()) {
      res.push_back(st.size());
      continue;
    }
    edges.erase(make_pair(x[i], y[i]));
    edges.erase(make_pair(y[i], x[i]));
    deg[x[i]]--;
    deg[y[i]]--;
    if (deg[x[i]] < k) remove(x[i]);
    if (st.find(y[i]) != st.end()) {
      if (deg[y[i]] < k) remove(y[i]);
    }
    res.push_back(st.size());
  }
  reverse(res.begin(), res.end());
  for (long long i = 0; i < res.size(); i++) {
    cout << res[i] << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<vector<long long> > adj;
set<long long> st;
vector<long long> deg;
long long k;
set<pair<long long, long long> > edges;
void remove(long long r) {
  st.erase(r);
  for (long long i = 0; i < adj[r].size(); i++) {
    long long c = adj[r][i];
    if (st.find(c) == st.end()) continue;
    if (edges.find(make_pair(r, c)) == edges.end()) continue;
    edges.erase(make_pair(r, c));
    edges.erase(make_pair(c, r));
    deg[c]--;
    if (deg[c] < k) remove(c);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m >> k;
  vector<long long> x(m);
  vector<long long> y(m);
  adj.resize(n);
  deg.resize(n, 0);
  for (long long i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    adj[x[i]].push_back(y[i]);
    adj[y[i]].push_back(x[i]);
    edges.insert(make_pair(x[i], y[i]));
    edges.insert(make_pair(y[i], x[i]));
    deg[x[i]]++;
    deg[y[i]]++;
  }
  for (long long i = 0; i < n; i++) {
    st.insert(i);
  }
  for (long long i = 0; i < n; i++) {
    if (st.find(i) == st.end()) continue;
    if (deg[i] < k) remove(i);
  }
  vector<long long> res;
  res.push_back(st.size());
  for (long long i = m - 1; i >= 1; i--) {
    if (st.find(x[i]) == st.end() || st.find(y[i]) == st.end()) {
      res.push_back(st.size());
      continue;
    }
    edges.erase(make_pair(x[i], y[i]));
    edges.erase(make_pair(y[i], x[i]));
    deg[x[i]]--;
    deg[y[i]]--;
    if (deg[x[i]] < k) remove(x[i]);
    if (st.find(y[i]) != st.end()) {
      if (deg[y[i]] < k) remove(y[i]);
    }
    res.push_back(st.size());
  }
  reverse(res.begin(), res.end());
  for (long long i = 0; i < res.size(); i++) {
    cout << res[i] << endl;
  }
  return 0;
}

```