#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
set<int> e[N];
int deg[N], x[N], y[N], ans[N], q[N], del[N];
int n, m, k, tail = 0, head = 1;
void delv(int u) {
  if (deg[u] >= k || del[u]) return;
  del[u] = 1;
  q[++tail] = u;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", x + i, y + i);
    e[x[i]].insert(y[i]);
    e[y[i]].insert(x[i]);
    ++deg[x[i]], ++deg[y[i]];
  }
  for (int i = 1; i <= n; ++i) delv(i);
  for (int i = m; i >= 1; --i) {
    while (head <= tail) {
      int u = q[head++];
      for (auto &v : e[u]) {
        --deg[v];
        delv(v);
        e[v].erase(u);
      }
    }
    ans[i] = n - tail;
    if (!del[x[i]] && !del[y[i]]) {
      --deg[x[i]], --deg[y[i]];
      delv(x[i]), delv(y[i]);
      e[x[i]].erase(y[i]);
      e[y[i]].erase(x[i]);
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d%c", ans[i], " \n"[i == m]);
  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 = 200005;
set<int> e[N];
int deg[N], x[N], y[N], ans[N], q[N], del[N];
int n, m, k, tail = 0, head = 1;
void delv(int u) {
  if (deg[u] >= k || del[u]) return;
  del[u] = 1;
  q[++tail] = u;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", x + i, y + i);
    e[x[i]].insert(y[i]);
    e[y[i]].insert(x[i]);
    ++deg[x[i]], ++deg[y[i]];
  }
  for (int i = 1; i <= n; ++i) delv(i);
  for (int i = m; i >= 1; --i) {
    while (head <= tail) {
      int u = q[head++];
      for (auto &v : e[u]) {
        --deg[v];
        delv(v);
        e[v].erase(u);
      }
    }
    ans[i] = n - tail;
    if (!del[x[i]] && !del[y[i]]) {
      --deg[x[i]], --deg[y[i]];
      delv(x[i]), delv(y[i]);
      e[x[i]].erase(y[i]);
      e[y[i]].erase(x[i]);
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d%c", ans[i], " \n"[i == m]);
  return 0;
}

```