Trips

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 $k$ ($2 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$, $1 \le k < n$) — 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 ($1 \leq i \leq m$) of the next $m$ lines contains two integers $x$ and $y$ ($1\leq x, y\leq n$, $x\ne y$), 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 ($1\leq i\leq m$) 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 $7$.

Solution:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m, k;
cin >> n >> m >> k;
vector<vector<pair<int,int>>> g(n);
vector<int> from(m), to(m);
for (int i = 0; i < m; i++) {
    cin >> from[i] >> to[i];
from[i]--; to[i]--;
g[from[i]].emplace_back(to[i], i);
g[to[i]].emplace_back(from[i], i);
}
vector<int> alive(m, 1);
vector<int> deg(n);
vector<int> que;
for (int i = 0; i < n; i++) {
    deg[i] = (int) g[i].size();
    if (deg[i] < k) {
      que.push_back(i);
    }
  }
  int b = 0;
  vector<int> res(m);
for (int step = m - 1; step >= 0; step--) {
while (b < (int) que.size()) {
      int i = que[b];
      for (auto &p : g[i]) {
        if (!alive[p.second]) {
          continue;
        }
        alive[p.second] = 0;
        if (deg[p.first] == k) {
          que.push_back(p.first);
        }
        deg[p.first]--;
        deg[i]--;
      }
      b++;
    }
    res[step] = n - (int) que.size();
    if (alive[step]) {
      if (deg[from[step]] == k) {
        que.push_back(from[step]);
      }
      if (deg[to[step]] == k) {
        que.push_back(to[step]);
      }
      deg[from[step]]--;
      deg[to[step]]--;
      alive[step] = 0;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << res[i] << '\n';
  }
  return 0;
}