#include <bits/stdc++.h>
using namespace std;
int deg[200005];
vector<pair<int, int> > g[200005];
bool notInSet[200005];
int main() {
  int n, m, k, x, y, i;
  cin >> n >> m >> k;
  vector<pair<int, int> > v;
  for (i = 0; i < m; i++) {
    cin >> x >> y;
    v.push_back({x, y});
    deg[x]++, deg[y]++;
    g[x].push_back({y, i});
    g[y].push_back({x, i});
  }
  set<pair<int, int> > s;
  for (i = 1; i <= n; i++) {
    s.insert({deg[i], i});
  }
  while (!s.empty() && deg[s.begin()->second] < k) {
    int u = s.begin()->second;
    for (auto it : g[u]) {
      int x = it.first;
      if (!notInSet[x]) {
        s.erase(s.find({deg[x], x}));
        deg[x]--;
        s.insert({deg[x], x});
      }
    }
    s.erase({deg[u], u});
    notInSet[u] = 1;
  }
  vector<int> res(m);
  for (i = m - 1; i >= 0; i--) {
    res[i] = s.size();
    x = v[i].first, y = v[i].second;
    if (!notInSet[x] && !notInSet[y]) {
      s.erase(s.find({deg[x], x}));
      s.erase(s.find({deg[y], y}));
      deg[x]--, deg[y]--;
      s.insert({deg[y], y});
      s.insert({deg[x], x});
      while (!s.empty() && deg[s.begin()->second] < k) {
        int u = s.begin()->second;
        for (auto it : g[u]) {
          int x = it.first;
          int y = it.second;
          if (y >= i) continue;
          if (!notInSet[x]) {
            s.erase(s.find({deg[x], x}));
            deg[x]--;
            s.insert({deg[x], x});
          }
        }
        s.erase({deg[u], u});
        notInSet[u] = 1;
      }
    }
  }
  for (i = 0; i < m; i++) cout << res[i] << endl;
}

### 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;
int deg[200005];
vector<pair<int, int> > g[200005];
bool notInSet[200005];
int main() {
  int n, m, k, x, y, i;
  cin >> n >> m >> k;
  vector<pair<int, int> > v;
  for (i = 0; i < m; i++) {
    cin >> x >> y;
    v.push_back({x, y});
    deg[x]++, deg[y]++;
    g[x].push_back({y, i});
    g[y].push_back({x, i});
  }
  set<pair<int, int> > s;
  for (i = 1; i <= n; i++) {
    s.insert({deg[i], i});
  }
  while (!s.empty() && deg[s.begin()->second] < k) {
    int u = s.begin()->second;
    for (auto it : g[u]) {
      int x = it.first;
      if (!notInSet[x]) {
        s.erase(s.find({deg[x], x}));
        deg[x]--;
        s.insert({deg[x], x});
      }
    }
    s.erase({deg[u], u});
    notInSet[u] = 1;
  }
  vector<int> res(m);
  for (i = m - 1; i >= 0; i--) {
    res[i] = s.size();
    x = v[i].first, y = v[i].second;
    if (!notInSet[x] && !notInSet[y]) {
      s.erase(s.find({deg[x], x}));
      s.erase(s.find({deg[y], y}));
      deg[x]--, deg[y]--;
      s.insert({deg[y], y});
      s.insert({deg[x], x});
      while (!s.empty() && deg[s.begin()->second] < k) {
        int u = s.begin()->second;
        for (auto it : g[u]) {
          int x = it.first;
          int y = it.second;
          if (y >= i) continue;
          if (!notInSet[x]) {
            s.erase(s.find({deg[x], x}));
            deg[x]--;
            s.insert({deg[x], x});
          }
        }
        s.erase({deg[u], u});
        notInSet[u] = 1;
      }
    }
  }
  for (i = 0; i < m; i++) cout << res[i] << endl;
}

```