#include <bits/stdc++.h>
using namespace std;
long long n, m, k, r[200005], redd[200005], t, xx[200005], yy[200005],
    ans[200005];
vector<pair<long long, long long> > v[200005];
void red(long long x, long long time) {
  if (time == 5) {
  }
  redd[x] = 1;
  t++;
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i].first > time) break;
    r[v[x][i].second]--;
    if (r[v[x][i].second] < k && redd[v[x][i].second] == 0) {
      red(v[x][i].second, time);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> xx[i] >> yy[i];
    v[xx[i]].push_back(make_pair(i, yy[i]));
    r[xx[i]]++;
    v[yy[i]].push_back(make_pair(i, xx[i]));
    r[yy[i]]++;
  }
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
  ans[m] = n;
  for (int i = 1; i <= n; i++) {
    if (r[i] < k && redd[i] == 0) {
      t = 0;
      red(i, m);
      ans[m] -= t;
    }
  }
  for (int i = m; i >= 2; i--) {
    for (int j = 1; j <= n; j++) {
    }
    if (redd[yy[i]] == 0) r[xx[i]]--;
    t = 0;
    if (r[xx[i]] < k && redd[xx[i]] == 0) {
      red(xx[i], i);
      ans[i - 1] = ans[i] - t;
      continue;
    }
    if (redd[xx[i]] == 0) r[yy[i]]--;
    if (r[yy[i]] < k && redd[yy[i]] == 0) {
      if (xx[i] == 5) {
      }
      r[xx[i]]++;
      red(yy[i], i);
    }
    ans[i - 1] = ans[i] - t;
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp to 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;
long long n, m, k, r[200005], redd[200005], t, xx[200005], yy[200005],
    ans[200005];
vector<pair<long long, long long> > v[200005];
void red(long long x, long long time) {
  if (time == 5) {
  }
  redd[x] = 1;
  t++;
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i].first > time) break;
    r[v[x][i].second]--;
    if (r[v[x][i].second] < k && redd[v[x][i].second] == 0) {
      red(v[x][i].second, time);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> xx[i] >> yy[i];
    v[xx[i]].push_back(make_pair(i, yy[i]));
    r[xx[i]]++;
    v[yy[i]].push_back(make_pair(i, xx[i]));
    r[yy[i]]++;
  }
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
  ans[m] = n;
  for (int i = 1; i <= n; i++) {
    if (r[i] < k && redd[i] == 0) {
      t = 0;
      red(i, m);
      ans[m] -= t;
    }
  }
  for (int i = m; i >= 2; i--) {
    for (int j = 1; j <= n; j++) {
    }
    if (redd[yy[i]] == 0) r[xx[i]]--;
    t = 0;
    if (r[xx[i]] < k && redd[xx[i]] == 0) {
      red(xx[i], i);
      ans[i - 1] = ans[i] - t;
      continue;
    }
    if (redd[xx[i]] == 0) r[yy[i]]--;
    if (r[yy[i]] < k && redd[yy[i]] == 0) {
      if (xx[i] == 5) {
      }
      r[xx[i]]++;
      red(yy[i], i);
    }
    ans[i - 1] = ans[i] - t;
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}

```