#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int du[maxn];
set<int> g[maxn];
int b1[maxn], b2[maxn], ans[maxn], vis[maxn];
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  set<pair<int, int> > s;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    b1[i] = x, b2[i] = y;
    du[x]++, du[y]++;
    g[x].insert(y);
    g[y].insert(x);
  }
  for (int i = 1; i <= n; i++) s.insert(make_pair(du[i], i));
  for (int i = m - 1; i >= 0; i--) {
    while (s.size() >= 1) {
      pair<int, int> itt = (*s.begin());
      if (itt.first >= k) break;
      for (set<int>::iterator it = g[itt.second].begin();
           it != g[itt.second].end(); it++) {
        int u = *it;
        if (vis[u] == 1) continue;
        s.erase(make_pair(du[u], u));
        du[u]--;
        s.insert(make_pair(du[u], u));
        g[u].erase(itt.second);
      }
      s.erase(itt);
      vis[itt.second] = 1;
    }
    ans[i] = s.size();
    if (vis[b1[i]] == 0 && vis[b2[i]] == 0) {
      s.erase(make_pair(du[b1[i]], b1[i]));
      du[b1[i]]--;
      s.insert(make_pair(du[b1[i]], b1[i]));
      g[b1[i]].erase(b2[i]);
      s.erase(make_pair(du[b2[i]], b2[i]));
      du[b2[i]]--;
      s.insert(make_pair(du[b2[i]], b2[i]));
      g[b2[i]].erase(b1[i]);
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}

### Prompt

Create a solution in Cpp for 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;
const int maxn = 2e5 + 10;
int du[maxn];
set<int> g[maxn];
int b1[maxn], b2[maxn], ans[maxn], vis[maxn];
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  set<pair<int, int> > s;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    b1[i] = x, b2[i] = y;
    du[x]++, du[y]++;
    g[x].insert(y);
    g[y].insert(x);
  }
  for (int i = 1; i <= n; i++) s.insert(make_pair(du[i], i));
  for (int i = m - 1; i >= 0; i--) {
    while (s.size() >= 1) {
      pair<int, int> itt = (*s.begin());
      if (itt.first >= k) break;
      for (set<int>::iterator it = g[itt.second].begin();
           it != g[itt.second].end(); it++) {
        int u = *it;
        if (vis[u] == 1) continue;
        s.erase(make_pair(du[u], u));
        du[u]--;
        s.insert(make_pair(du[u], u));
        g[u].erase(itt.second);
      }
      s.erase(itt);
      vis[itt.second] = 1;
    }
    ans[i] = s.size();
    if (vis[b1[i]] == 0 && vis[b2[i]] == 0) {
      s.erase(make_pair(du[b1[i]], b1[i]));
      du[b1[i]]--;
      s.insert(make_pair(du[b1[i]], b1[i]));
      g[b1[i]].erase(b2[i]);
      s.erase(make_pair(du[b2[i]], b2[i]));
      du[b2[i]]--;
      s.insert(make_pair(du[b2[i]], b2[i]));
      g[b2[i]].erase(b1[i]);
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}

```