#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 200;
int n, m, k, id[N], vis[N], ans[N];
vector<pair<int, int> > g[N];
struct node {
  int x, y;
} e[N];
set<pair<int, int> > s;
void change(int x) {
  if (vis[x]) return;
  auto it = s.lower_bound(make_pair(id[x], x));
  pair<int, int> t = *it;
  s.erase(it);
  t.first--;
  id[t.second]--;
  s.insert(t);
}
void update(int now) {
  while (!s.empty() && (*s.begin()).first < k) {
    pair<int, int> t = *s.begin();
    vis[t.second] = 1;
    s.erase(s.begin());
    for (auto pp : g[t.second]) {
      if (pp.second > now) break;
      int v = pp.first;
      if (vis[v]) continue;
      auto it = s.lower_bound(make_pair(id[v], v));
      t = *it;
      s.erase(it);
      id[v]--;
      t.first--;
      s.insert(t);
    }
  }
}
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> e[i].x >> e[i].y;
    g[e[i].x].push_back(make_pair(e[i].y, i));
    g[e[i].y].push_back(make_pair(e[i].x, i));
    id[e[i].x]++;
    id[e[i].y]++;
  }
  for (int i = 1; i <= n; i++) sort(g[i].begin(), g[i].end(), cmp);
  for (int i = 1; i <= n; i++) s.insert(make_pair(id[i], i));
  for (int i = m; i >= 1; i--) {
    update(i);
    ans[i] = s.size();
    if (vis[e[i].x] || vis[e[i].y]) continue;
    int l = id[e[i].x], r = id[e[i].y];
    change(e[i].x);
    change(e[i].y);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int N = 2e5 + 200;
int n, m, k, id[N], vis[N], ans[N];
vector<pair<int, int> > g[N];
struct node {
  int x, y;
} e[N];
set<pair<int, int> > s;
void change(int x) {
  if (vis[x]) return;
  auto it = s.lower_bound(make_pair(id[x], x));
  pair<int, int> t = *it;
  s.erase(it);
  t.first--;
  id[t.second]--;
  s.insert(t);
}
void update(int now) {
  while (!s.empty() && (*s.begin()).first < k) {
    pair<int, int> t = *s.begin();
    vis[t.second] = 1;
    s.erase(s.begin());
    for (auto pp : g[t.second]) {
      if (pp.second > now) break;
      int v = pp.first;
      if (vis[v]) continue;
      auto it = s.lower_bound(make_pair(id[v], v));
      t = *it;
      s.erase(it);
      id[v]--;
      t.first--;
      s.insert(t);
    }
  }
}
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> e[i].x >> e[i].y;
    g[e[i].x].push_back(make_pair(e[i].y, i));
    g[e[i].y].push_back(make_pair(e[i].x, i));
    id[e[i].x]++;
    id[e[i].y]++;
  }
  for (int i = 1; i <= n; i++) sort(g[i].begin(), g[i].end(), cmp);
  for (int i = 1; i <= n; i++) s.insert(make_pair(id[i], i));
  for (int i = m; i >= 1; i--) {
    update(i);
    ans[i] = s.size();
    if (vis[e[i].x] || vis[e[i].y]) continue;
    int l = id[e[i].x], r = id[e[i].y];
    change(e[i].x);
    change(e[i].y);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  return 0;
}

```