#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
int ans[N];
int deg[N];
bool trip[N];
set<int> g[N];
int cnt, k;
void remove(int u) {
  if (!trip[u]) return;
  queue<int> q;
  q.push(u);
  trip[u] = false;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    cnt--;
    for (int v : g[u]) {
      deg[v]--;
      g[v].erase(u);
      if (trip[v] and deg[v] < k) {
        trip[v] = false;
        q.push(v);
      }
    }
    g[u].clear();
  }
}
int main() {
  int n, m;
  scanf("%d %d %d", &n, &m, &k);
  vector<pair<int, int> > e;
  cnt = n;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    deg[x]++;
    deg[y]++;
    e.emplace_back(x, y);
    g[x].insert(y);
    g[y].insert(x);
  }
  for (int i = 1; i <= n; i++) trip[i] = true;
  for (int i = 1; i <= n; i++) {
    if (trip[i] and deg[i] < k) remove(i);
  }
  ans[m - 1] = cnt;
  for (int i = m - 2; i >= 0; i--) {
    int a = e[i + 1].first, b = e[i + 1].second;
    if (!trip[a] or !trip[b]) {
      ans[i] = cnt;
      continue;
    }
    deg[a]--;
    deg[b]--;
    g[a].erase(b);
    g[b].erase(a);
    if (deg[a] < k) remove(a);
    if (deg[b] < k) remove(b);
    ans[i] = cnt;
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 = 200010;
int ans[N];
int deg[N];
bool trip[N];
set<int> g[N];
int cnt, k;
void remove(int u) {
  if (!trip[u]) return;
  queue<int> q;
  q.push(u);
  trip[u] = false;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    cnt--;
    for (int v : g[u]) {
      deg[v]--;
      g[v].erase(u);
      if (trip[v] and deg[v] < k) {
        trip[v] = false;
        q.push(v);
      }
    }
    g[u].clear();
  }
}
int main() {
  int n, m;
  scanf("%d %d %d", &n, &m, &k);
  vector<pair<int, int> > e;
  cnt = n;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    deg[x]++;
    deg[y]++;
    e.emplace_back(x, y);
    g[x].insert(y);
    g[y].insert(x);
  }
  for (int i = 1; i <= n; i++) trip[i] = true;
  for (int i = 1; i <= n; i++) {
    if (trip[i] and deg[i] < k) remove(i);
  }
  ans[m - 1] = cnt;
  for (int i = m - 2; i >= 0; i--) {
    int a = e[i + 1].first, b = e[i + 1].second;
    if (!trip[a] or !trip[b]) {
      ans[i] = cnt;
      continue;
    }
    deg[a]--;
    deg[b]--;
    g[a].erase(b);
    g[b].erase(a);
    if (deg[a] < k) remove(a);
    if (deg[b] < k) remove(b);
    ans[i] = cnt;
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

```