#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int n, m, k, used[N], w[N], all, deg[N];
bool c[N];
vector<set<int> > g(N);
vector<int> res;
vector<pair<int, int> > reb;
queue<int> q;
void bfs() {
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (auto& u : g[v]) {
      if (deg[u] == k) {
        deg[u]--;
        q.push(u);
      } else
        deg[u]--;
    }
  }
}
void ddfs(int v) {
  for (auto& u : g[v]) {
    if (w[u] == 1) {
      w[u] = 0;
      c[u] = 0;
      all--;
      ddfs(u);
    } else
      w[u]--;
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].insert(b);
    g[b].insert(a);
    deg[a]++;
    deg[b]++;
    reb.push_back({a, b});
  }
  for (int i = 0; i < n; i++) {
    if (deg[i] < k) {
      q.push(i);
    }
  }
  bfs();
  for (int i = 0; i < n; i++)
    if (deg[i] >= k) c[i] = 1;
  for (int i = 0; i < n; i++) {
    int x = 0;
    for (auto& v : g[i]) x += c[v];
    w[i] = max(0, x - k + 1);
    all += c[i];
  }
  res.push_back(all);
  for (int i = m - 1; i >= 0; i--) {
    int a = reb[i].first, b = reb[i].second;
    if (c[a] && c[b]) {
      if (w[a] == 1) {
        w[a] = 0;
        all--;
        c[a] = 0;
        ddfs(a);
      } else if (w[b] == 1) {
        w[b] = 0;
        all--;
        c[b] = 0;
        ddfs(b);
      } else {
        w[a]--;
        w[b]--;
      }
    }
    g[a].erase(b);
    g[b].erase(a);
    res.push_back(all);
  }
  for (int i = m - 1; i >= 0; i--) cout << res[i] << '\n';
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int n, m, k, used[N], w[N], all, deg[N];
bool c[N];
vector<set<int> > g(N);
vector<int> res;
vector<pair<int, int> > reb;
queue<int> q;
void bfs() {
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (auto& u : g[v]) {
      if (deg[u] == k) {
        deg[u]--;
        q.push(u);
      } else
        deg[u]--;
    }
  }
}
void ddfs(int v) {
  for (auto& u : g[v]) {
    if (w[u] == 1) {
      w[u] = 0;
      c[u] = 0;
      all--;
      ddfs(u);
    } else
      w[u]--;
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].insert(b);
    g[b].insert(a);
    deg[a]++;
    deg[b]++;
    reb.push_back({a, b});
  }
  for (int i = 0; i < n; i++) {
    if (deg[i] < k) {
      q.push(i);
    }
  }
  bfs();
  for (int i = 0; i < n; i++)
    if (deg[i] >= k) c[i] = 1;
  for (int i = 0; i < n; i++) {
    int x = 0;
    for (auto& v : g[i]) x += c[v];
    w[i] = max(0, x - k + 1);
    all += c[i];
  }
  res.push_back(all);
  for (int i = m - 1; i >= 0; i--) {
    int a = reb[i].first, b = reb[i].second;
    if (c[a] && c[b]) {
      if (w[a] == 1) {
        w[a] = 0;
        all--;
        c[a] = 0;
        ddfs(a);
      } else if (w[b] == 1) {
        w[b] = 0;
        all--;
        c[b] = 0;
        ddfs(b);
      } else {
        w[a]--;
        w[b]--;
      }
    }
    g[a].erase(b);
    g[b].erase(a);
    res.push_back(all);
  }
  for (int i = m - 1; i >= 0; i--) cout << res[i] << '\n';
}

```