#include <bits/stdc++.h>
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << " = " << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ", ";
  _dbg(sdbg + 1, a...);
}
const int maxn = (1e6) + 7;
const int maxk = 20;
const int inf = (1e9) + 7;
const long long LLinf = (1e18) + 7;
const long double eps = 1e-9;
const long long mod = 1e9 + 7;
queue<int> q;
bool vis[maxn];
int res;
int n, m, k;
int odp[maxn];
set<int> wek[maxn];
pair<int, int> tab[maxn];
void rob() {
  while (((int)(q).size())) {
    int a = q.front();
    q.pop();
    if (vis[a] == 0) continue;
    vis[a] = 0;
    res--;
    for (auto s : wek[a])
      if (vis[s] == 1) {
        wek[s].erase(a);
        if (((int)(wek[s]).size()) < k) q.push(s);
      }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  res = n;
  for (int i = 1; i <= m; i++) {
    cin >> tab[i].first >> tab[i].second;
    wek[tab[i].first].insert(tab[i].second);
    wek[tab[i].second].insert(tab[i].first);
  }
  for (int i = 1; i <= n; i++) {
    vis[i] = 1;
    if (((int)(wek[i]).size()) < k) q.push(i);
  }
  rob();
  odp[m] = res;
  for (int i = m - 1; i >= 0; i--) {
    wek[tab[i + 1].first].erase(tab[i + 1].second);
    wek[tab[i + 1].second].erase(tab[i + 1].first);
    if (vis[tab[i + 1].first] == 1 && ((int)(wek[tab[i + 1].first]).size()) < k)
      q.push(tab[i + 1].first);
    if (vis[tab[i + 1].second] == 1 &&
        ((int)(wek[tab[i + 1].second]).size()) < k)
      q.push(tab[i + 1].second);
    rob();
    odp[i] = res;
  }
  for (int i = 1; i <= m; i++) cout << odp[i] << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << " = " << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ", ";
  _dbg(sdbg + 1, a...);
}
const int maxn = (1e6) + 7;
const int maxk = 20;
const int inf = (1e9) + 7;
const long long LLinf = (1e18) + 7;
const long double eps = 1e-9;
const long long mod = 1e9 + 7;
queue<int> q;
bool vis[maxn];
int res;
int n, m, k;
int odp[maxn];
set<int> wek[maxn];
pair<int, int> tab[maxn];
void rob() {
  while (((int)(q).size())) {
    int a = q.front();
    q.pop();
    if (vis[a] == 0) continue;
    vis[a] = 0;
    res--;
    for (auto s : wek[a])
      if (vis[s] == 1) {
        wek[s].erase(a);
        if (((int)(wek[s]).size()) < k) q.push(s);
      }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  res = n;
  for (int i = 1; i <= m; i++) {
    cin >> tab[i].first >> tab[i].second;
    wek[tab[i].first].insert(tab[i].second);
    wek[tab[i].second].insert(tab[i].first);
  }
  for (int i = 1; i <= n; i++) {
    vis[i] = 1;
    if (((int)(wek[i]).size()) < k) q.push(i);
  }
  rob();
  odp[m] = res;
  for (int i = m - 1; i >= 0; i--) {
    wek[tab[i + 1].first].erase(tab[i + 1].second);
    wek[tab[i + 1].second].erase(tab[i + 1].first);
    if (vis[tab[i + 1].first] == 1 && ((int)(wek[tab[i + 1].first]).size()) < k)
      q.push(tab[i + 1].first);
    if (vis[tab[i + 1].second] == 1 &&
        ((int)(wek[tab[i + 1].second]).size()) < k)
      q.push(tab[i + 1].second);
    rob();
    odp[i] = res;
  }
  for (int i = 1; i <= m; i++) cout << odp[i] << "\n";
  return 0;
}

```