#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
vector<int> G[N];
int n, m, k;
int qu[N], qv[N], vis[N], du[N], ans[N], res;
void dfs(int u) {
  if (vis[u]) return;
  if (du[u] < k) {
    vis[u] = 1;
    for (int v : G[u])
      if (!vis[v]) du[v]--;
    for (int v : G[u]) {
      if (!vis[v]) {
        dfs(v);
      }
    }
    res--;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> qu[i] >> qv[i];
    G[qu[i]].push_back(qv[i]);
    G[qv[i]].push_back(qu[i]);
    du[qu[i]]++;
    du[qv[i]]++;
  }
  res = n;
  for (int i = 1; i <= n; i++) {
    dfs(i);
  }
  ans[m] = res;
  for (int i = m; i >= 1; i--) {
    if (!vis[qu[i]] && !vis[qv[i]]) {
      if (du[qu[i]] - 1 < k) {
        du[qu[i]]--;
        dfs(qu[i]);
      } else if (du[qv[i]] - 1 < k) {
        du[qv[i]]--;
        dfs(qv[i]);
      } else {
        du[qu[i]]--;
        du[qv[i]]--;
      }
    }
    G[qu[i]].pop_back();
    G[qv[i]].pop_back();
    ans[i - 1] = res;
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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 int N = 2e5 + 10;
vector<int> G[N];
int n, m, k;
int qu[N], qv[N], vis[N], du[N], ans[N], res;
void dfs(int u) {
  if (vis[u]) return;
  if (du[u] < k) {
    vis[u] = 1;
    for (int v : G[u])
      if (!vis[v]) du[v]--;
    for (int v : G[u]) {
      if (!vis[v]) {
        dfs(v);
      }
    }
    res--;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> qu[i] >> qv[i];
    G[qu[i]].push_back(qv[i]);
    G[qv[i]].push_back(qu[i]);
    du[qu[i]]++;
    du[qv[i]]++;
  }
  res = n;
  for (int i = 1; i <= n; i++) {
    dfs(i);
  }
  ans[m] = res;
  for (int i = m; i >= 1; i--) {
    if (!vis[qu[i]] && !vis[qv[i]]) {
      if (du[qu[i]] - 1 < k) {
        du[qu[i]]--;
        dfs(qu[i]);
      } else if (du[qv[i]] - 1 < k) {
        du[qv[i]]--;
        dfs(qv[i]);
      } else {
        du[qu[i]]--;
        du[qv[i]]--;
      }
    }
    G[qu[i]].pop_back();
    G[qv[i]].pop_back();
    ans[i - 1] = res;
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}

```