#include <bits/stdc++.h>
using namespace std;
void del(int edge, vector<set<int>>& g, map<int, int>& degr, int k) {
  degr.erase(degr.find(edge));
  for (auto neigh : g[edge]) {
    g[neigh].erase(edge);
    auto it = degr.find(neigh);
    if (it != degr.end()) {
      if (--degr[neigh] < k) {
        del(neigh, g, degr, k);
      }
    }
  }
  g[edge].clear();
}
int main() {
  int n, m, k;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> k;
  vector<set<int>> g(n);
  vector<pair<int, int>> edges;
  map<int, int> degr;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].insert(b);
    g[b].insert(a);
    edges.push_back({a, b});
    degr[a]++;
    degr[b]++;
  }
  for (int i = 0; i < n; i++) {
    if (degr.find(i) != degr.end() && degr[i] < k) {
      del(i, g, degr, k);
    }
  }
  std::list<int> answ;
  for (int i = m - 1; i >= 0; i--) {
    answ.push_front(degr.size());
    auto edge = edges[i];
    if (g[edge.second].find(edge.first) != g[edge.second].end()) {
      g[edge.second].erase(edge.first);
      if (--degr[edge.first] < k) del(edge.first, g, degr, k);
    }
    if (g[edge.first].find(edge.second) != g[edge.first].end()) {
      g[edge.first].erase(edge.second);
      if (--degr[edge.second] < k) del(edge.second, g, degr, k);
    }
  }
  for (auto a : answ) {
    cout << a << "\n";
  }
  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;
void del(int edge, vector<set<int>>& g, map<int, int>& degr, int k) {
  degr.erase(degr.find(edge));
  for (auto neigh : g[edge]) {
    g[neigh].erase(edge);
    auto it = degr.find(neigh);
    if (it != degr.end()) {
      if (--degr[neigh] < k) {
        del(neigh, g, degr, k);
      }
    }
  }
  g[edge].clear();
}
int main() {
  int n, m, k;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> k;
  vector<set<int>> g(n);
  vector<pair<int, int>> edges;
  map<int, int> degr;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].insert(b);
    g[b].insert(a);
    edges.push_back({a, b});
    degr[a]++;
    degr[b]++;
  }
  for (int i = 0; i < n; i++) {
    if (degr.find(i) != degr.end() && degr[i] < k) {
      del(i, g, degr, k);
    }
  }
  std::list<int> answ;
  for (int i = m - 1; i >= 0; i--) {
    answ.push_front(degr.size());
    auto edge = edges[i];
    if (g[edge.second].find(edge.first) != g[edge.second].end()) {
      g[edge.second].erase(edge.first);
      if (--degr[edge.first] < k) del(edge.first, g, degr, k);
    }
    if (g[edge.first].find(edge.second) != g[edge.first].end()) {
      g[edge.first].erase(edge.second);
      if (--degr[edge.second] < k) del(edge.second, g, degr, k);
    }
  }
  for (auto a : answ) {
    cout << a << "\n";
  }
  return 0;
}

```