#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
using namespace std;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m, k;
  cin >> n >> m >> k;
  vector<set<long long>> g(n + 1);
  vector<long long> deg(n + 1, 0);
  vector<pair<long long, long long>> v(m);
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    g[a].insert(b);
    g[b].insert(a);
    deg[a]++;
    deg[b]++;
    v[i].first = a;
    v[i].second = b;
  }
  long long ans = n;
  queue<long long> q;
  vector<long long> vis(n + 1, 0);
  for (long long i = 1; i <= n; i++) {
    if (deg[i] < k) {
      q.push(i);
      vis[i] = 1;
    }
  }
  while (!q.empty()) {
    long long d = q.front();
    q.pop();
    ans--;
    for (auto it : g[d]) {
      g[it].erase(d);
      deg[it]--;
      if (deg[it] == k - 1) q.push(it);
    }
    g[d].clear();
  }
  vector<long long> fans(m);
  for (long long i = m - 1; i >= 0; i--) {
    fans[i] = ans;
    long long a = v[i].first;
    long long b = v[i].second;
    if (g[a].find(b) != g[a].end()) {
      g[a].erase(b);
      deg[a]--;
      if (deg[a] == k - 1) q.push(a);
    }
    if (g[b].find(a) != g[b].end()) {
      g[b].erase(a);
      deg[b]--;
      if (deg[b] == k - 1) q.push(b);
    }
    while (!q.empty()) {
      long long d = q.front();
      q.pop();
      ans--;
      for (auto it : g[d]) {
        g[it].erase(d);
        deg[it]--;
        if (deg[it] == k - 1) q.push(it);
      }
      g[d].clear();
    }
  }
  for (long long i = 0; i < m; i++) {
    cout << fans[i] << "\n";
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#pragma comment(linker, "/stack:200000000")
using namespace std;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m, k;
  cin >> n >> m >> k;
  vector<set<long long>> g(n + 1);
  vector<long long> deg(n + 1, 0);
  vector<pair<long long, long long>> v(m);
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    g[a].insert(b);
    g[b].insert(a);
    deg[a]++;
    deg[b]++;
    v[i].first = a;
    v[i].second = b;
  }
  long long ans = n;
  queue<long long> q;
  vector<long long> vis(n + 1, 0);
  for (long long i = 1; i <= n; i++) {
    if (deg[i] < k) {
      q.push(i);
      vis[i] = 1;
    }
  }
  while (!q.empty()) {
    long long d = q.front();
    q.pop();
    ans--;
    for (auto it : g[d]) {
      g[it].erase(d);
      deg[it]--;
      if (deg[it] == k - 1) q.push(it);
    }
    g[d].clear();
  }
  vector<long long> fans(m);
  for (long long i = m - 1; i >= 0; i--) {
    fans[i] = ans;
    long long a = v[i].first;
    long long b = v[i].second;
    if (g[a].find(b) != g[a].end()) {
      g[a].erase(b);
      deg[a]--;
      if (deg[a] == k - 1) q.push(a);
    }
    if (g[b].find(a) != g[b].end()) {
      g[b].erase(a);
      deg[b]--;
      if (deg[b] == k - 1) q.push(b);
    }
    while (!q.empty()) {
      long long d = q.front();
      q.pop();
      ans--;
      for (auto it : g[d]) {
        g[it].erase(d);
        deg[it]--;
        if (deg[it] == k - 1) q.push(it);
      }
      g[d].clear();
    }
  }
  for (long long i = 0; i < m; i++) {
    cout << fans[i] << "\n";
  }
  return 0;
}

```