#include <bits/stdc++.h>
using namespace std;
int n, m, k;
set<int> M[200005];
int siz[200005], a[200005], b[200005], dele[200005], cnt, ans[200005];
void check(int u) {
  if (siz[u] >= k || dele[u]) return;
  cnt--;
  dele[u] = 1;
  queue<int> Q;
  Q.push(u);
  while (!Q.empty()) {
    int ty = Q.front();
    Q.pop();
    for (set<int>::iterator i = M[ty].begin(); i != M[ty].end(); i++) {
      int v = *i;
      siz[v]--;
      if (siz[v] < k && !dele[v]) {
        cnt--;
        Q.push(v);
        dele[v] = 1;
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i], &b[i]);
    M[a[i]].insert(b[i]);
    M[b[i]].insert(a[i]);
    siz[a[i]]++;
    siz[b[i]]++;
  }
  cnt = n;
  for (int i = 1; i <= n; i++) check(i);
  ans[m] = cnt;
  for (int i = m; i >= 2; i--) {
    if (!dele[b[i]]) siz[a[i]]--;
    if (!dele[a[i]]) siz[b[i]]--;
    M[a[i]].erase(b[i]);
    M[b[i]].erase(a[i]);
    check(a[i]);
    check(b[i]);
    ans[i - 1] = cnt;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  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;
int n, m, k;
set<int> M[200005];
int siz[200005], a[200005], b[200005], dele[200005], cnt, ans[200005];
void check(int u) {
  if (siz[u] >= k || dele[u]) return;
  cnt--;
  dele[u] = 1;
  queue<int> Q;
  Q.push(u);
  while (!Q.empty()) {
    int ty = Q.front();
    Q.pop();
    for (set<int>::iterator i = M[ty].begin(); i != M[ty].end(); i++) {
      int v = *i;
      siz[v]--;
      if (siz[v] < k && !dele[v]) {
        cnt--;
        Q.push(v);
        dele[v] = 1;
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i], &b[i]);
    M[a[i]].insert(b[i]);
    M[b[i]].insert(a[i]);
    siz[a[i]]++;
    siz[b[i]]++;
  }
  cnt = n;
  for (int i = 1; i <= n; i++) check(i);
  ans[m] = cnt;
  for (int i = m; i >= 2; i--) {
    if (!dele[b[i]]) siz[a[i]]--;
    if (!dele[a[i]]) siz[b[i]]--;
    M[a[i]].erase(b[i]);
    M[b[i]].erase(a[i]);
    check(a[i]);
    check(b[i]);
    ans[i - 1] = cnt;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```