#include <bits/stdc++.h>
using namespace std;
const long long N = 201102;
vector<long long> g[N];
set<pair<long long, long long> > s;
long long n, m, k, v, u, d[N], a[N], b[N], ans[N];
pair<long long, long long> x;
set<pair<long long, long long> >::iterator it;
map<long long, bool> e[N];
int main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  for (long long i = 1; i <= m; i++) {
    scanf("%lld%lld", &v, &u);
    d[v]++, d[u]++;
    g[v].push_back(u), g[u].push_back(v);
    a[i] = v, b[i] = u;
  }
  for (long long i = 1; i <= n; i++) s.insert({d[i], i});
  for (long long i = m + 1; i >= 1 and s.size(); i--) {
    v = a[i];
    if (e[a[i]][b[i]]) {
      ans[i] = s.size();
      continue;
    }
    e[b[i]][a[i]] = e[a[i]][b[i]] = 1;
    if (d[v] >= k) {
      s.erase({d[v], v});
      d[v]--;
      s.insert({d[v], v});
    }
    v = b[i];
    if (d[v] >= k) {
      s.erase({d[v], v});
      d[v]--;
      s.insert({d[v], v});
    }
    while (s.size()) {
      v = (*s.begin()).second;
      if (d[v] >= k) break;
      s.erase({d[v], v});
      d[v] = 0;
      for (long long j = 0; j < g[v].size(); j++) {
        u = g[v][j];
        if (d[u] < k or e[u][v] or s.find({d[u], u}) == s.end()) continue;
        e[u][v] = e[v][u] = 1;
        s.erase({d[u], u});
        d[u]--;
        s.insert({d[u], u});
      }
    }
    ans[i] = s.size();
  }
  for (long long i = 2; i <= m + 1; i++) printf("%lld\n", ans[i]);
}

### 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 long long N = 201102;
vector<long long> g[N];
set<pair<long long, long long> > s;
long long n, m, k, v, u, d[N], a[N], b[N], ans[N];
pair<long long, long long> x;
set<pair<long long, long long> >::iterator it;
map<long long, bool> e[N];
int main() {
  scanf("%lld%lld%lld", &n, &m, &k);
  for (long long i = 1; i <= m; i++) {
    scanf("%lld%lld", &v, &u);
    d[v]++, d[u]++;
    g[v].push_back(u), g[u].push_back(v);
    a[i] = v, b[i] = u;
  }
  for (long long i = 1; i <= n; i++) s.insert({d[i], i});
  for (long long i = m + 1; i >= 1 and s.size(); i--) {
    v = a[i];
    if (e[a[i]][b[i]]) {
      ans[i] = s.size();
      continue;
    }
    e[b[i]][a[i]] = e[a[i]][b[i]] = 1;
    if (d[v] >= k) {
      s.erase({d[v], v});
      d[v]--;
      s.insert({d[v], v});
    }
    v = b[i];
    if (d[v] >= k) {
      s.erase({d[v], v});
      d[v]--;
      s.insert({d[v], v});
    }
    while (s.size()) {
      v = (*s.begin()).second;
      if (d[v] >= k) break;
      s.erase({d[v], v});
      d[v] = 0;
      for (long long j = 0; j < g[v].size(); j++) {
        u = g[v][j];
        if (d[u] < k or e[u][v] or s.find({d[u], u}) == s.end()) continue;
        e[u][v] = e[v][u] = 1;
        s.erase({d[u], u});
        d[u]--;
        s.insert({d[u], u});
      }
    }
    ans[i] = s.size();
  }
  for (long long i = 2; i <= m + 1; i++) printf("%lld\n", ans[i]);
}

```