#include <bits/stdc++.h>
using namespace std;
const int mx = 2e5 + 10, cut = 700;
int n, m, k, cnt[mx], lv, x[mx], y[mx], dead[mx];
vector<pair<int, int> > adj[mx];
vector<int> res;
void er(int a, int b) {
  auto it = lower_bound(adj[a].begin(), adj[a].end(), make_pair(b, 0));
  if (it->second) return;
  it->second = 1;
  cnt[a]--;
}
void f(int h) {
  if (cnt[h] >= k || dead[h]) return;
  dead[h] = 1;
  lv--;
  for (auto &it : adj[h])
    if (!it.second) er(it.first, h);
  for (auto &it : adj[h])
    if (!it.second) f(it.first);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  if (k > cut) {
    for (int i = 0; i < m; i++) puts("0");
    return 0;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d", x + i, y + i);
    cnt[x[i]]++;
    cnt[y[i]]++;
    adj[x[i]].push_back({y[i], 0});
    adj[y[i]].push_back({x[i], 0});
  }
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  lv = n;
  for (int i = 1; i <= n; i++) f(i);
  for (int i = m; i--;) {
    res.push_back(lv);
    er(x[i], y[i]);
    er(y[i], x[i]);
    f(x[i]);
    f(y[i]);
  }
  reverse(res.begin(), res.end());
  for (auto &it : res) printf("%d\n", it);
  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 mx = 2e5 + 10, cut = 700;
int n, m, k, cnt[mx], lv, x[mx], y[mx], dead[mx];
vector<pair<int, int> > adj[mx];
vector<int> res;
void er(int a, int b) {
  auto it = lower_bound(adj[a].begin(), adj[a].end(), make_pair(b, 0));
  if (it->second) return;
  it->second = 1;
  cnt[a]--;
}
void f(int h) {
  if (cnt[h] >= k || dead[h]) return;
  dead[h] = 1;
  lv--;
  for (auto &it : adj[h])
    if (!it.second) er(it.first, h);
  for (auto &it : adj[h])
    if (!it.second) f(it.first);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  if (k > cut) {
    for (int i = 0; i < m; i++) puts("0");
    return 0;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d", x + i, y + i);
    cnt[x[i]]++;
    cnt[y[i]]++;
    adj[x[i]].push_back({y[i], 0});
    adj[y[i]].push_back({x[i], 0});
  }
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  lv = n;
  for (int i = 1; i <= n; i++) f(i);
  for (int i = m; i--;) {
    res.push_back(lv);
    er(x[i], y[i]);
    er(y[i], x[i]);
    f(x[i]);
    f(y[i]);
  }
  reverse(res.begin(), res.end());
  for (auto &it : res) printf("%d\n", it);
  return 0;
}

```