#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int n, m, k, indi[(int)5e5 + 110];
set<int> gra[(int)5e5 + 110];
int arr[(int)5e5 + 110], bra[(int)5e5 + 110], result[(int)5e5 + 110],
    donno[(int)5e5 + 110];
int track;
void funck(int u) {
  queue<int> pq;
  pq.push(u);
  while (!pq.empty()) {
    u = pq.front();
    pq.pop();
    if (donno[u]) continue;
    donno[u] = 1;
    track--;
    for (int go : gra[u]) {
      gra[go].erase(u);
      if (--indi[go] < k) pq.push(go);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    gra[u].insert(v);
    gra[v].insert(u);
    indi[u]++;
    indi[v]++;
    arr[i] = u;
    bra[i] = v;
  }
  track = n;
  for (int i = 0; i < n; ++i) {
    if (donno[i] == 0 and indi[i] < k) {
      funck(i);
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    result[i] = track;
    if (donno[arr[i]] == 0 and donno[bra[i]] == 0) {
      indi[arr[i]]--;
      indi[bra[i]]--;
      gra[arr[i]].erase(bra[i]);
      gra[bra[i]].erase(arr[i]);
      if (indi[arr[i]] < k) funck(arr[i]);
      if (indi[bra[i]] < k) funck(bra[i]);
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << result[i] << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 = 1000000007;
int n, m, k, indi[(int)5e5 + 110];
set<int> gra[(int)5e5 + 110];
int arr[(int)5e5 + 110], bra[(int)5e5 + 110], result[(int)5e5 + 110],
    donno[(int)5e5 + 110];
int track;
void funck(int u) {
  queue<int> pq;
  pq.push(u);
  while (!pq.empty()) {
    u = pq.front();
    pq.pop();
    if (donno[u]) continue;
    donno[u] = 1;
    track--;
    for (int go : gra[u]) {
      gra[go].erase(u);
      if (--indi[go] < k) pq.push(go);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    gra[u].insert(v);
    gra[v].insert(u);
    indi[u]++;
    indi[v]++;
    arr[i] = u;
    bra[i] = v;
  }
  track = n;
  for (int i = 0; i < n; ++i) {
    if (donno[i] == 0 and indi[i] < k) {
      funck(i);
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    result[i] = track;
    if (donno[arr[i]] == 0 and donno[bra[i]] == 0) {
      indi[arr[i]]--;
      indi[bra[i]]--;
      gra[arr[i]].erase(bra[i]);
      gra[bra[i]].erase(arr[i]);
      if (indi[arr[i]] < k) funck(arr[i]);
      if (indi[bra[i]] < k) funck(bra[i]);
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << result[i] << endl;
  }
  return 0;
}

```