#include <bits/stdc++.h>
using namespace std;
int intcmp(const void *v1, const void *v2) { return *(int *)v1 - *(int *)v2; }
int n, m, k;
vector<pair<int, int>> v1v2index[200010];
int ea[200010], eb[200010];
int d[200010];
int curidx;
bool deleted[200010];
int delcnt[200010];
int delnow;
void dfs(int v) {
  if (deleted[v] || d[v] >= k) {
    return;
  }
  deleted[v] = true;
  ++delnow;
  for (auto [u, idx] : v1v2index[v]) {
    if (idx >= curidx) {
      continue;
    }
    --d[u];
    dfs(u);
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    v1v2index[a].push_back(make_pair(b, i));
    v1v2index[b].push_back(make_pair(a, i));
    ea[i] = a;
    eb[i] = b;
    d[a]++;
    d[b]++;
  }
  curidx = m;
  for (int i = 1; i < n + 1; i++) {
    dfs(i);
  }
  for (curidx = m - 1; curidx >= 0; curidx--) {
    delcnt[curidx] = delnow;
    if (deleted[ea[curidx]] || deleted[eb[curidx]]) {
      continue;
    }
    d[ea[curidx]]--;
    d[eb[curidx]]--;
    dfs(ea[curidx]);
    dfs(eb[curidx]);
  }
  for (int i = 0; i < m; ++i) {
    printf("%d\n", n - delcnt[i]);
  }
  return 0;
}

### Prompt

Please create a solution in cpp to 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 intcmp(const void *v1, const void *v2) { return *(int *)v1 - *(int *)v2; }
int n, m, k;
vector<pair<int, int>> v1v2index[200010];
int ea[200010], eb[200010];
int d[200010];
int curidx;
bool deleted[200010];
int delcnt[200010];
int delnow;
void dfs(int v) {
  if (deleted[v] || d[v] >= k) {
    return;
  }
  deleted[v] = true;
  ++delnow;
  for (auto [u, idx] : v1v2index[v]) {
    if (idx >= curidx) {
      continue;
    }
    --d[u];
    dfs(u);
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    v1v2index[a].push_back(make_pair(b, i));
    v1v2index[b].push_back(make_pair(a, i));
    ea[i] = a;
    eb[i] = b;
    d[a]++;
    d[b]++;
  }
  curidx = m;
  for (int i = 1; i < n + 1; i++) {
    dfs(i);
  }
  for (curidx = m - 1; curidx >= 0; curidx--) {
    delcnt[curidx] = delnow;
    if (deleted[ea[curidx]] || deleted[eb[curidx]]) {
      continue;
    }
    d[ea[curidx]]--;
    d[eb[curidx]]--;
    dfs(ea[curidx]);
    dfs(eb[curidx]);
  }
  for (int i = 0; i < m; ++i) {
    printf("%d\n", n - delcnt[i]);
  }
  return 0;
}

```