#include <bits/stdc++.h>
using namespace std;
const long long N = 5e5;
const long long INF = 1e18;
const long long mod = 1e9 + 7;
set<long long> g[N], st;
long long i, j, n, m, k, ans[N], u[N], v[N];
void dfs(long long x) {
  auto it = st.lower_bound(x);
  if (*it != x || g[x].size() >= k) return;
  st.erase(it);
  for (auto to : g[x]) {
    g[to].erase(x);
    dfs(to);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (i = 0; i < m; ++i) {
    cin >> u[i] >> v[i];
    g[u[i]].insert(v[i]);
    g[v[i]].insert(u[i]);
  }
  for (i = 1; i <= n; ++i) st.insert(i);
  for (i = 1; i <= n; ++i) dfs(i);
  for (i = m - 1; i >= 0; --i) {
    long long x = u[i], y = v[i];
    ans[i] = st.size();
    g[x].erase(y);
    g[y].erase(x);
    dfs(x);
    dfs(y);
  }
  for (i = 0; i < m; ++i) cout << ans[i] << "\n";
}

### 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;
const long long N = 5e5;
const long long INF = 1e18;
const long long mod = 1e9 + 7;
set<long long> g[N], st;
long long i, j, n, m, k, ans[N], u[N], v[N];
void dfs(long long x) {
  auto it = st.lower_bound(x);
  if (*it != x || g[x].size() >= k) return;
  st.erase(it);
  for (auto to : g[x]) {
    g[to].erase(x);
    dfs(to);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (i = 0; i < m; ++i) {
    cin >> u[i] >> v[i];
    g[u[i]].insert(v[i]);
    g[v[i]].insert(u[i]);
  }
  for (i = 1; i <= n; ++i) st.insert(i);
  for (i = 1; i <= n; ++i) dfs(i);
  for (i = m - 1; i >= 0; --i) {
    long long x = u[i], y = v[i];
    ans[i] = st.size();
    g[x].erase(y);
    g[y].erase(x);
    dfs(x);
    dfs(y);
  }
  for (i = 0; i < m; ++i) cout << ans[i] << "\n";
}

```