#include <bits/stdc++.h>
const int N = 200005;
int n, m, k, ans[N], in[N], cnt;
std::map<int, bool> vis[N];
std::vector<int> G[N];
std::vector<std::pair<int, int> > e;
std::queue<int> q;
void topo() {
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    in[u] = 0;
    for (int i = 0; i < (signed)G[u].size(); i++) {
      int v = G[u][i];
      if (vis[u][v]) continue;
      vis[u][v] = vis[v][u] = 1, in[v]--;
      if (in[v] == k - 1) cnt--, q.push(v);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    e.push_back(std::make_pair(x, y));
    G[x].push_back(y), G[y].push_back(x);
    in[x]++, in[y]++;
  }
  cnt = n;
  for (int i = 1; i <= n; i++)
    if (in[i] < k) cnt--, q.push(i);
  topo(), ans[m] = cnt;
  for (int i = m; i >= 2; i--) {
    int u = e[i - 1].first, v = e[i - 1].second;
    if (vis[u][v]) {
      ans[i - 1] = ans[i];
      continue;
    }
    vis[u][v] = vis[v][u] = 1;
    in[u]--, in[v]--;
    if (in[u] == k - 1) cnt--, q.push(u);
    if (in[v] == k - 1) cnt--, q.push(v);
    topo(), ans[i - 1] = cnt;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### 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>
const int N = 200005;
int n, m, k, ans[N], in[N], cnt;
std::map<int, bool> vis[N];
std::vector<int> G[N];
std::vector<std::pair<int, int> > e;
std::queue<int> q;
void topo() {
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    in[u] = 0;
    for (int i = 0; i < (signed)G[u].size(); i++) {
      int v = G[u][i];
      if (vis[u][v]) continue;
      vis[u][v] = vis[v][u] = 1, in[v]--;
      if (in[v] == k - 1) cnt--, q.push(v);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    e.push_back(std::make_pair(x, y));
    G[x].push_back(y), G[y].push_back(x);
    in[x]++, in[y]++;
  }
  cnt = n;
  for (int i = 1; i <= n; i++)
    if (in[i] < k) cnt--, q.push(i);
  topo(), ans[m] = cnt;
  for (int i = m; i >= 2; i--) {
    int u = e[i - 1].first, v = e[i - 1].second;
    if (vis[u][v]) {
      ans[i - 1] = ans[i];
      continue;
    }
    vis[u][v] = vis[v][u] = 1;
    in[u]--, in[v]--;
    if (in[u] == k - 1) cnt--, q.push(u);
    if (in[v] == k - 1) cnt--, q.push(v);
    topo(), ans[i - 1] = cnt;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```