#include <bits/stdc++.h>
using namespace std;
set<long long> V[400010];
long long n, m, k, D[400010], A[400010], B[400010], ans;
signed main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  ans = n;
  for (long long i = 1; i <= m; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    A[i] = a;
    B[i] = b;
    D[a]++;
    D[b]++;
    V[a].insert(b);
    V[b].insert(a);
  }
  vector<long long> Tp, Ans;
  Ans.clear();
  Tp.clear();
  for (long long i = 1; i <= n; i++)
    if (D[i] < k) Tp.push_back(i);
  for (long long i = m; i >= 1; i--) {
    for (long long j = 0; j < Tp.size(); j++) {
      long long x = Tp[j];
      if (D[x] <= -100000)
        continue;
      else
        ans--;
      for (set<long long>::iterator it = V[x].begin(); it != V[x].end(); it++) {
        V[*it].erase(x);
        if (--D[*it] < k) Tp.push_back(*it);
      }
      V[x].clear();
      D[x] = -100000;
    }
    Ans.push_back(ans);
    Tp.clear();
    if (!V[A[i]].count(B[i])) continue;
    D[A[i]]--;
    D[B[i]]--;
    V[A[i]].erase(B[i]);
    V[B[i]].erase(A[i]);
    if (D[A[i]] < k) Tp.push_back(A[i]);
    if (D[B[i]] < k) Tp.push_back(B[i]);
  }
  for (long long i = (long long)Ans.size() - 1; i >= 0; i--)
    printf("%lld\n", Ans[i]);
  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;
set<long long> V[400010];
long long n, m, k, D[400010], A[400010], B[400010], ans;
signed main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  ans = n;
  for (long long i = 1; i <= m; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    A[i] = a;
    B[i] = b;
    D[a]++;
    D[b]++;
    V[a].insert(b);
    V[b].insert(a);
  }
  vector<long long> Tp, Ans;
  Ans.clear();
  Tp.clear();
  for (long long i = 1; i <= n; i++)
    if (D[i] < k) Tp.push_back(i);
  for (long long i = m; i >= 1; i--) {
    for (long long j = 0; j < Tp.size(); j++) {
      long long x = Tp[j];
      if (D[x] <= -100000)
        continue;
      else
        ans--;
      for (set<long long>::iterator it = V[x].begin(); it != V[x].end(); it++) {
        V[*it].erase(x);
        if (--D[*it] < k) Tp.push_back(*it);
      }
      V[x].clear();
      D[x] = -100000;
    }
    Ans.push_back(ans);
    Tp.clear();
    if (!V[A[i]].count(B[i])) continue;
    D[A[i]]--;
    D[B[i]]--;
    V[A[i]].erase(B[i]);
    V[B[i]].erase(A[i]);
    if (D[A[i]] < k) Tp.push_back(A[i]);
    if (D[B[i]] < k) Tp.push_back(B[i]);
  }
  for (long long i = (long long)Ans.size() - 1; i >= 0; i--)
    printf("%lld\n", Ans[i]);
  return 0;
}

```