#include <bits/stdc++.h>
using namespace std;
long long M = 2e5 + 10;
vector<long long> v[200009], deg(M), vis(M);
map<pair<long long, long long>, long long> ed;
void dfs(long long s, long long k, long long p, long long &ans) {
  vis[s] = 1;
  ans--;
  for (auto u : v[s]) {
    pair<long long, long long> p1;
    p1.first = u;
    p1.second = s;
    if (u != p && ed.find(p1) == ed.end()) {
      deg[u]--;
      deg[s]--;
      ed[make_pair(u, s)] = 1;
      ed[make_pair(s, u)] = 1;
    }
    if (vis[u] == 0 && deg[u] < k) {
      dfs(u, k, s, ans);
    }
  }
  return;
}
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<pair<long long, long long> > edge;
  for (long long i = 0; i < m; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
    edge.push_back(make_pair(a, b));
    deg[a]++;
    deg[b]++;
  }
  long long ans = n;
  vector<long long> ans1(m);
  vis[0] = 1;
  for (long long i = 1; i <= n; i++) {
    if (deg[i] < k && vis[i] == 0) {
      dfs(i, k, 0, ans);
    }
  }
  for (long long i = m - 1; i >= 0; i--) {
    ans1[i] = ans;
    long long a = edge[i].first;
    long long b = edge[i].second;
    if (vis[a] == 0 && vis[b] == 0 && ed.find(make_pair(a, b)) == ed.end()) {
      deg[a]--;
      deg[b]--;
      ed[make_pair(a, b)] = 1;
      ed[make_pair(b, a)] = 1;
    }
    if (deg[a] < k && vis[a] == 0) {
      dfs(a, k, b, ans);
    }
    if (deg[b] < k && vis[b] == 0) {
      dfs(b, k, a, ans);
    }
  }
  for (long long i = 0; i < m; i++) {
    printf("%lld\n", ans1[i]);
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to 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;
long long M = 2e5 + 10;
vector<long long> v[200009], deg(M), vis(M);
map<pair<long long, long long>, long long> ed;
void dfs(long long s, long long k, long long p, long long &ans) {
  vis[s] = 1;
  ans--;
  for (auto u : v[s]) {
    pair<long long, long long> p1;
    p1.first = u;
    p1.second = s;
    if (u != p && ed.find(p1) == ed.end()) {
      deg[u]--;
      deg[s]--;
      ed[make_pair(u, s)] = 1;
      ed[make_pair(s, u)] = 1;
    }
    if (vis[u] == 0 && deg[u] < k) {
      dfs(u, k, s, ans);
    }
  }
  return;
}
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<pair<long long, long long> > edge;
  for (long long i = 0; i < m; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
    edge.push_back(make_pair(a, b));
    deg[a]++;
    deg[b]++;
  }
  long long ans = n;
  vector<long long> ans1(m);
  vis[0] = 1;
  for (long long i = 1; i <= n; i++) {
    if (deg[i] < k && vis[i] == 0) {
      dfs(i, k, 0, ans);
    }
  }
  for (long long i = m - 1; i >= 0; i--) {
    ans1[i] = ans;
    long long a = edge[i].first;
    long long b = edge[i].second;
    if (vis[a] == 0 && vis[b] == 0 && ed.find(make_pair(a, b)) == ed.end()) {
      deg[a]--;
      deg[b]--;
      ed[make_pair(a, b)] = 1;
      ed[make_pair(b, a)] = 1;
    }
    if (deg[a] < k && vis[a] == 0) {
      dfs(a, k, b, ans);
    }
    if (deg[b] < k && vis[b] == 0) {
      dfs(b, k, a, ans);
    }
  }
  for (long long i = 0; i < m; i++) {
    printf("%lld\n", ans1[i]);
  }
  return 0;
}

```