#include <bits/stdc++.h>
using namespace std;
int n, m, k;
int x[202020];
int y[202020];
set<int> v[202020];
int d[202020];
int z[202020];
vector<int> q;
int c;
vector<int> r;
int p[202020];
void lol() {
  for (int i = 0; i < q.size(); i++) {
    int x = q[i];
    if (!z[x]) continue;
    z[x] = 0;
    c--;
    for (auto y : v[x]) {
      d[y]--;
      if (d[y] < k) {
        if (p[y]) continue;
        p[y] = 1;
        q.push_back(y);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) z[i] = 1;
  for (int i = 1; i <= m; i++) {
    cin >> x[i] >> y[i];
    v[x[i]].insert(y[i]);
    d[x[i]]++;
    v[y[i]].insert(x[i]);
    d[y[i]]++;
  }
  c = n;
  for (int i = 1; i <= n; i++) {
    if (d[i] < k) q.push_back(i);
  }
  lol();
  r.push_back(c);
  for (int i = m; i >= 1; i--) {
    q.clear();
    if (z[x[i]] && z[y[i]]) {
      v[x[i]].erase(y[i]);
      v[y[i]].erase(x[i]);
      d[x[i]]--;
      d[y[i]]--;
      if (d[x[i]] < k) q.push_back(x[i]);
      if (d[y[i]] < k) q.push_back(y[i]);
      lol();
    }
    if (i != 1) r.push_back(c);
  }
  reverse(r.begin(), r.end());
  for (auto x : r) cout << x << "\n";
}

### 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;
int x[202020];
int y[202020];
set<int> v[202020];
int d[202020];
int z[202020];
vector<int> q;
int c;
vector<int> r;
int p[202020];
void lol() {
  for (int i = 0; i < q.size(); i++) {
    int x = q[i];
    if (!z[x]) continue;
    z[x] = 0;
    c--;
    for (auto y : v[x]) {
      d[y]--;
      if (d[y] < k) {
        if (p[y]) continue;
        p[y] = 1;
        q.push_back(y);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) z[i] = 1;
  for (int i = 1; i <= m; i++) {
    cin >> x[i] >> y[i];
    v[x[i]].insert(y[i]);
    d[x[i]]++;
    v[y[i]].insert(x[i]);
    d[y[i]]++;
  }
  c = n;
  for (int i = 1; i <= n; i++) {
    if (d[i] < k) q.push_back(i);
  }
  lol();
  r.push_back(c);
  for (int i = m; i >= 1; i--) {
    q.clear();
    if (z[x[i]] && z[y[i]]) {
      v[x[i]].erase(y[i]);
      v[y[i]].erase(x[i]);
      d[x[i]]--;
      d[y[i]]--;
      if (d[x[i]] < k) q.push_back(x[i]);
      if (d[y[i]] < k) q.push_back(y[i]);
      lol();
    }
    if (i != 1) r.push_back(c);
  }
  reverse(r.begin(), r.end());
  for (auto x : r) cout << x << "\n";
}

```