#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
vector<vector<pair<int, int> > > g;
set<pair<int, int> > good;
vector<char> very_good;
vector<int> d;
int k;
void check(long long int i) {
  while (!good.empty() && good.begin()->first < k) {
    int v = good.begin()->second;
    for (auto &y : g[v]) {
      int x = y.first;
      if (y.second >= i) continue;
      if (very_good[x]) {
        good.erase(make_pair(d[x], x));
        d[x]--;
        good.insert(make_pair(d[x], x));
      }
    }
    good.erase(make_pair(d[v], v));
    very_good[v] = false;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m >> k;
  g.resize(n);
  d.resize(n);
  vector<pair<int, int> > f(m);
  for (int i = 0; i < m; ++i) {
    cin >> f[i].first >> f[i].second;
    f[i].first--, f[i].second--;
    g[f[i].first].push_back(make_pair(f[i].second, i));
    g[f[i].second].push_back(make_pair(f[i].first, i));
    d[f[i].first]++;
    d[f[i].second]++;
  }
  for (int i = 0; i < n; ++i) {
    good.insert(make_pair(d[i], i));
  }
  very_good.assign(n, true);
  check(1e18);
  vector<int> ans(m);
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = good.size();
    int v = f[i].first, u = f[i].second;
    if (very_good[v] && very_good[u]) {
      good.erase(make_pair(d[v], v));
      d[v]--;
      good.insert(make_pair(d[v], v));
      good.erase(make_pair(d[u], u));
      d[u]--;
      good.insert(make_pair(d[u], u));
      check(i);
    }
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << '\n';
}

### Prompt

Develop a solution in cpp 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>
using namespace std;
const int mod = 1e9 + 7;
vector<vector<pair<int, int> > > g;
set<pair<int, int> > good;
vector<char> very_good;
vector<int> d;
int k;
void check(long long int i) {
  while (!good.empty() && good.begin()->first < k) {
    int v = good.begin()->second;
    for (auto &y : g[v]) {
      int x = y.first;
      if (y.second >= i) continue;
      if (very_good[x]) {
        good.erase(make_pair(d[x], x));
        d[x]--;
        good.insert(make_pair(d[x], x));
      }
    }
    good.erase(make_pair(d[v], v));
    very_good[v] = false;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m >> k;
  g.resize(n);
  d.resize(n);
  vector<pair<int, int> > f(m);
  for (int i = 0; i < m; ++i) {
    cin >> f[i].first >> f[i].second;
    f[i].first--, f[i].second--;
    g[f[i].first].push_back(make_pair(f[i].second, i));
    g[f[i].second].push_back(make_pair(f[i].first, i));
    d[f[i].first]++;
    d[f[i].second]++;
  }
  for (int i = 0; i < n; ++i) {
    good.insert(make_pair(d[i], i));
  }
  very_good.assign(n, true);
  check(1e18);
  vector<int> ans(m);
  for (int i = m - 1; i >= 0; --i) {
    ans[i] = good.size();
    int v = f[i].first, u = f[i].second;
    if (very_good[v] && very_good[u]) {
      good.erase(make_pair(d[v], v));
      d[v]--;
      good.insert(make_pair(d[v], v));
      good.erase(make_pair(d[u], u));
      d[u]--;
      good.insert(make_pair(d[u], u));
      check(i);
    }
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << '\n';
}

```