#include <bits/stdc++.h>
using namespace std;
const long long linf = 1LL << 62;
const int iinf = 1000000009;
const double dinf = 1e17;
const int Mod = 1e9 + 9;
const int maxn = 1000005;
set<pair<int, int> > s, sp;
int n, m, k, u, v;
int mp[maxn], rt;
pair<int, int> sd[maxn], ds[maxn];
vector<int> vp[200005];
void up(int x) {
  for (int i = 1; i <= vp[x].size(); i++) {
    if (s.count(make_pair(vp[x][i - 1], x))) {
      sp.erase(make_pair(mp[vp[x][i - 1]], vp[x][i - 1]));
      mp[vp[x][i - 1]]--;
      sp.insert(make_pair(mp[vp[x][i - 1]], vp[x][i - 1]));
      s.erase(make_pair(vp[x][i - 1], x));
      s.erase(make_pair(x, vp[x][i - 1]));
    }
  }
}
void del() {
  set<pair<int, int> >::iterator it;
  while (1) {
    rt = 0;
    for (it = sp.begin(); it != sp.end(); it++) {
      if (((*it).first) >= k) break;
      ds[rt++] = (*it);
    }
    if (!rt) break;
    for (int i = 0; i <= rt - 1; i++) sp.erase(ds[i]), up(ds[i].second);
  }
}
void del(int i) {
  if (s.count(sd[i])) {
    s.erase(sd[i]);
    u = sd[i].first;
    v = sd[i].second;
    s.erase(make_pair(v, u));
    sp.erase(make_pair(mp[u], u));
    sp.insert(make_pair(--mp[u], u));
    sp.erase(make_pair(mp[v], v));
    sp.insert(make_pair(--mp[v], v));
    del();
  }
}
int ans[maxn];
void solve() {
  s.clear();
  ios::sync_with_stdio(false);
  memset(mp, 0, sizeof(mp));
  sp.clear();
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) vp[i].clear();
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    mp[u]++;
    mp[v]++;
    s.insert(make_pair(u, v));
    s.insert(make_pair(v, u));
    sd[i] = make_pair(u, v);
    vp[v].push_back(u);
    vp[u].push_back(v);
  }
  for (int i = 1; i <= n; i++) sp.insert(make_pair(mp[i], i));
  for (int i = m; i >= 1; i--) {
    del();
    ans[i] = sp.size();
    del(i);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  ;
}
int main() {
  int t = 1;
  while (t--) solve();
  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;
const long long linf = 1LL << 62;
const int iinf = 1000000009;
const double dinf = 1e17;
const int Mod = 1e9 + 9;
const int maxn = 1000005;
set<pair<int, int> > s, sp;
int n, m, k, u, v;
int mp[maxn], rt;
pair<int, int> sd[maxn], ds[maxn];
vector<int> vp[200005];
void up(int x) {
  for (int i = 1; i <= vp[x].size(); i++) {
    if (s.count(make_pair(vp[x][i - 1], x))) {
      sp.erase(make_pair(mp[vp[x][i - 1]], vp[x][i - 1]));
      mp[vp[x][i - 1]]--;
      sp.insert(make_pair(mp[vp[x][i - 1]], vp[x][i - 1]));
      s.erase(make_pair(vp[x][i - 1], x));
      s.erase(make_pair(x, vp[x][i - 1]));
    }
  }
}
void del() {
  set<pair<int, int> >::iterator it;
  while (1) {
    rt = 0;
    for (it = sp.begin(); it != sp.end(); it++) {
      if (((*it).first) >= k) break;
      ds[rt++] = (*it);
    }
    if (!rt) break;
    for (int i = 0; i <= rt - 1; i++) sp.erase(ds[i]), up(ds[i].second);
  }
}
void del(int i) {
  if (s.count(sd[i])) {
    s.erase(sd[i]);
    u = sd[i].first;
    v = sd[i].second;
    s.erase(make_pair(v, u));
    sp.erase(make_pair(mp[u], u));
    sp.insert(make_pair(--mp[u], u));
    sp.erase(make_pair(mp[v], v));
    sp.insert(make_pair(--mp[v], v));
    del();
  }
}
int ans[maxn];
void solve() {
  s.clear();
  ios::sync_with_stdio(false);
  memset(mp, 0, sizeof(mp));
  sp.clear();
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) vp[i].clear();
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    mp[u]++;
    mp[v]++;
    s.insert(make_pair(u, v));
    s.insert(make_pair(v, u));
    sd[i] = make_pair(u, v);
    vp[v].push_back(u);
    vp[u].push_back(v);
  }
  for (int i = 1; i <= n; i++) sp.insert(make_pair(mp[i], i));
  for (int i = m; i >= 1; i--) {
    del();
    ans[i] = sp.size();
    del(i);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  ;
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}

```