#include <bits/stdc++.h>
using namespace std;
int n, m, k, cnt[222222], x[222222], y[222222], all, idx[222222], idy[222222],
    ans[222222];
bool used[222222];
vector<pair<int, int> > g[222222];
void upd(int i) {
  used[i] = 1;
  all--;
  for (int j = 0; j < g[i].size(); j++) {
    if (!g[i][j].second) continue;
    cnt[g[i][j].first]--;
    if (used[g[i][j].first]) continue;
    if (cnt[g[i][j].first] < k) upd(g[i][j].first);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x[i], &y[i]);
    idx[i] = g[x[i]].size();
    idy[i] = g[y[i]].size();
    g[x[i]].push_back(make_pair(y[i], 1));
    g[y[i]].push_back(make_pair(x[i], 1));
  }
  all = n;
  for (int i = 1; i <= n; i++) cnt[i] = g[i].size();
  for (int i = 1; i <= n; i++) {
    if (!used[i] && cnt[i] < k) upd(i);
  }
  ans[m] = all;
  for (int i = m; i >= 2; i--) {
    g[x[i]][idx[i]].second = 0;
    g[y[i]][idy[i]].second = 0;
    if (used[x[i]] | used[y[i]]) {
      ans[i - 1] = all;
      continue;
    }
    cnt[x[i]]--;
    cnt[y[i]]--;
    if (cnt[x[i]] < k) upd(x[i]);
    if (!used[y[i]] && cnt[y[i]] < k) upd(y[i]);
    ans[i - 1] = all;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
int n, m, k, cnt[222222], x[222222], y[222222], all, idx[222222], idy[222222],
    ans[222222];
bool used[222222];
vector<pair<int, int> > g[222222];
void upd(int i) {
  used[i] = 1;
  all--;
  for (int j = 0; j < g[i].size(); j++) {
    if (!g[i][j].second) continue;
    cnt[g[i][j].first]--;
    if (used[g[i][j].first]) continue;
    if (cnt[g[i][j].first] < k) upd(g[i][j].first);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x[i], &y[i]);
    idx[i] = g[x[i]].size();
    idy[i] = g[y[i]].size();
    g[x[i]].push_back(make_pair(y[i], 1));
    g[y[i]].push_back(make_pair(x[i], 1));
  }
  all = n;
  for (int i = 1; i <= n; i++) cnt[i] = g[i].size();
  for (int i = 1; i <= n; i++) {
    if (!used[i] && cnt[i] < k) upd(i);
  }
  ans[m] = all;
  for (int i = m; i >= 2; i--) {
    g[x[i]][idx[i]].second = 0;
    g[y[i]][idy[i]].second = 0;
    if (used[x[i]] | used[y[i]]) {
      ans[i - 1] = all;
      continue;
    }
    cnt[x[i]]--;
    cnt[y[i]]--;
    if (cnt[x[i]] < k) upd(x[i]);
    if (!used[y[i]] && cnt[y[i]] < k) upd(y[i]);
    ans[i - 1] = all;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```