#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
unordered_set<int> e[N];
int n, m, k, ex[N], ey[N], i, ans, ot[N];
bool bb[N];
void del(int x) {
  bb[x] = 0;
  --ans;
  for (int y : e[x]) e[y].erase(x);
  for (int y : e[x])
    if (bb[y] && e[y].size() < k) del(y);
  e[x].clear();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (i = 1; i <= m; ++i)
    cin >> ex[i] >> ey[i], e[ex[i]].insert(ey[i]), e[ey[i]].insert(ex[i]);
  memset(bb + 1, 1, n);
  ans = n;
  for (i = 1; i <= n; ++i)
    if (bb[i] && e[i].size() < k) del(i);
  for (i = m; i; --i) {
    ot[i] = ans;
    if (bb[ex[i]] && bb[ey[i]]) {
      e[ex[i]].erase(ey[i]);
      e[ey[i]].erase(ex[i]);
      if (bb[ex[i]] && e[ex[i]].size() < k) del(ex[i]);
      if (bb[ey[i]] && e[ey[i]].size() < k) del(ey[i]);
    }
  }
  for (i = 1; i <= m; ++i) cout << ot[i] << '\n';
  return 0;
}

### Prompt

In cpp, your task is to solve 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 N = 2e5 + 5;
unordered_set<int> e[N];
int n, m, k, ex[N], ey[N], i, ans, ot[N];
bool bb[N];
void del(int x) {
  bb[x] = 0;
  --ans;
  for (int y : e[x]) e[y].erase(x);
  for (int y : e[x])
    if (bb[y] && e[y].size() < k) del(y);
  e[x].clear();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (i = 1; i <= m; ++i)
    cin >> ex[i] >> ey[i], e[ex[i]].insert(ey[i]), e[ey[i]].insert(ex[i]);
  memset(bb + 1, 1, n);
  ans = n;
  for (i = 1; i <= n; ++i)
    if (bb[i] && e[i].size() < k) del(i);
  for (i = m; i; --i) {
    ot[i] = ans;
    if (bb[ex[i]] && bb[ey[i]]) {
      e[ex[i]].erase(ey[i]);
      e[ey[i]].erase(ex[i]);
      if (bb[ex[i]] && e[ex[i]].size() < k) del(ex[i]);
      if (bb[ey[i]] && e[ey[i]].size() < k) del(ey[i]);
    }
  }
  for (i = 1; i <= m; ++i) cout << ot[i] << '\n';
  return 0;
}

```