#include <bits/stdc++.h>
using namespace std;
int n, m, k, numOn, res[200000];
set<int> f[200000];
bool on[200000];
vector<pair<int, int> > v;
queue<int> toRemove;
void update() {
  while (!toRemove.empty()) {
    int r = toRemove.front();
    toRemove.pop();
    numOn--;
    for (auto next : f[r]) {
      f[next].erase(r);
      if (on[next] && ((int)(f[next]).size()) < k) {
        on[next] = 0;
        toRemove.push(next);
      }
    }
    f[r].clear();
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < (m); i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    v.push_back(make_pair(x, y));
    f[x].insert(y), f[y].insert(x);
  }
  for (int i = 0; i < (n); i++) on[i] = 1;
  numOn = n;
  for (int i = 0; i < (n); i++)
    if (((int)(f[i]).size()) < k) {
      on[i] = 0;
      toRemove.push(i);
    }
  update();
  res[m - 1] = numOn;
  for (int i = (m)-1; i >= (1); i--) {
    int x = v[i].first, y = v[i].second;
    f[x].erase(y), f[y].erase(x);
    if (on[x] && ((int)(f[x]).size()) < k) {
      on[x] = 0;
      toRemove.push(x);
    }
    if (on[y] && ((int)(f[y]).size()) < k) {
      on[y] = 0;
      toRemove.push(y);
    }
    update();
    res[i - 1] = numOn;
  }
  for (int i = 0; i < (m); i++) cout << res[i] << "\n";
}

### 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 n, m, k, numOn, res[200000];
set<int> f[200000];
bool on[200000];
vector<pair<int, int> > v;
queue<int> toRemove;
void update() {
  while (!toRemove.empty()) {
    int r = toRemove.front();
    toRemove.pop();
    numOn--;
    for (auto next : f[r]) {
      f[next].erase(r);
      if (on[next] && ((int)(f[next]).size()) < k) {
        on[next] = 0;
        toRemove.push(next);
      }
    }
    f[r].clear();
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < (m); i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    v.push_back(make_pair(x, y));
    f[x].insert(y), f[y].insert(x);
  }
  for (int i = 0; i < (n); i++) on[i] = 1;
  numOn = n;
  for (int i = 0; i < (n); i++)
    if (((int)(f[i]).size()) < k) {
      on[i] = 0;
      toRemove.push(i);
    }
  update();
  res[m - 1] = numOn;
  for (int i = (m)-1; i >= (1); i--) {
    int x = v[i].first, y = v[i].second;
    f[x].erase(y), f[y].erase(x);
    if (on[x] && ((int)(f[x]).size()) < k) {
      on[x] = 0;
      toRemove.push(x);
    }
    if (on[y] && ((int)(f[y]).size()) < k) {
      on[y] = 0;
      toRemove.push(y);
    }
    update();
    res[i - 1] = numOn;
  }
  for (int i = 0; i < (m); i++) cout << res[i] << "\n";
}

```