#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n, m, k;
int du[maxn];
set<int> g[maxn];
vector<pair<int, int> > ar;
set<pair<int, int> > st;
set<int>::iterator it;
int ans[maxn];
void dfs() {
  if (st.size() == 0) return;
  if (st.begin()->first < k) {
    int u = st.begin()->second;
    st.erase(st.begin());
    for (it = g[u].begin(); it != g[u].end(); it++) {
      int v = *it;
      if (st.count(make_pair(du[v], v))) {
        st.erase(make_pair(du[v], v));
        du[v]--;
        st.insert(make_pair(du[v], v));
      }
    }
    dfs();
  }
}
int main() {
  int u, v;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    du[u]++;
    du[v]++;
    g[u].insert(v);
    g[v].insert(u);
    ar.push_back(make_pair(u, v));
  }
  for (int i = 1; i <= n; i++) {
    st.insert(make_pair(du[i], i));
  }
  dfs();
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = st.size();
    u = ar[i].first;
    v = ar[i].second;
    if (st.count(make_pair(du[u], u)) && st.count(make_pair(du[v], v))) {
      st.erase(make_pair(du[u], u));
      st.erase(make_pair(du[v], v));
      du[u]--;
      du[v]--;
      st.insert(make_pair(du[u], u));
      st.insert(make_pair(du[v], v));
      g[u].erase(v);
      g[v].erase(u);
      dfs();
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

### 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 int maxn = 2e5 + 5;
int n, m, k;
int du[maxn];
set<int> g[maxn];
vector<pair<int, int> > ar;
set<pair<int, int> > st;
set<int>::iterator it;
int ans[maxn];
void dfs() {
  if (st.size() == 0) return;
  if (st.begin()->first < k) {
    int u = st.begin()->second;
    st.erase(st.begin());
    for (it = g[u].begin(); it != g[u].end(); it++) {
      int v = *it;
      if (st.count(make_pair(du[v], v))) {
        st.erase(make_pair(du[v], v));
        du[v]--;
        st.insert(make_pair(du[v], v));
      }
    }
    dfs();
  }
}
int main() {
  int u, v;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    du[u]++;
    du[v]++;
    g[u].insert(v);
    g[v].insert(u);
    ar.push_back(make_pair(u, v));
  }
  for (int i = 1; i <= n; i++) {
    st.insert(make_pair(du[i], i));
  }
  dfs();
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = st.size();
    u = ar[i].first;
    v = ar[i].second;
    if (st.count(make_pair(du[u], u)) && st.count(make_pair(du[v], v))) {
      st.erase(make_pair(du[u], u));
      st.erase(make_pair(du[v], v));
      du[u]--;
      du[v]--;
      st.insert(make_pair(du[u], u));
      st.insert(make_pair(du[v], v));
      g[u].erase(v);
      g[v].erase(u);
      dfs();
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

```