#include <bits/stdc++.h>
using namespace std;
const int MAXN = 210000;
int n, m, k;
int s[MAXN], t[MAXN];
int d[MAXN];
int ans[MAXN], vis[MAXN];
vector<int> edge[MAXN];
pair<int, int> mk(int x, int y) {
  return x > y ? make_pair(y, x) : make_pair(x, y);
}
set<pair<int, int> > S;
queue<int> q;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &s[i], &t[i]);
    edge[s[i]].push_back(t[i]);
    edge[t[i]].push_back(s[i]);
    d[s[i]]++, d[t[i]]++;
  }
  int cnt = n;
  for (int i = 1; i <= n; i++) {
    if (d[i] < k) {
      q.push(i);
      vis[i] = 1;
      cnt--;
    }
  }
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (unsigned i = 0; i < edge[now].size(); i++) {
      int v = edge[now][i];
      if (S.count(mk(now, v))) continue;
      S.insert(mk(now, v));
      d[now]--, d[v]--;
      if (d[v] < k && vis[v] == 0) {
        q.push(v);
        vis[v] = 1;
        cnt--;
      }
    }
  }
  for (int i = m; i >= 1; --i) {
    ans[i] = cnt;
    if (vis[s[i]] || vis[t[i]] || S.count(mk(s[i], t[i]))) continue;
    S.insert(mk(s[i], t[i]));
    d[s[i]]--, d[t[i]]--;
    if (d[s[i]] < k && vis[s[i]] == 0) {
      q.push(s[i]);
      vis[s[i]] = 1;
      cnt--;
    }
    if (d[t[i]] < k && vis[t[i]] == 0) {
      q.push(t[i]);
      vis[t[i]] = 1;
      cnt--;
    }
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      for (unsigned i = 0; i < edge[now].size(); i++) {
        int v = edge[now][i];
        if (S.count(mk(now, v))) continue;
        d[now]--, d[v]--;
        if (d[v] < k && vis[v] == 0) {
          S.insert(mk(now, v));
          q.push(v);
          vis[v] = 1;
          cnt--;
        }
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Your task is to create 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 MAXN = 210000;
int n, m, k;
int s[MAXN], t[MAXN];
int d[MAXN];
int ans[MAXN], vis[MAXN];
vector<int> edge[MAXN];
pair<int, int> mk(int x, int y) {
  return x > y ? make_pair(y, x) : make_pair(x, y);
}
set<pair<int, int> > S;
queue<int> q;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &s[i], &t[i]);
    edge[s[i]].push_back(t[i]);
    edge[t[i]].push_back(s[i]);
    d[s[i]]++, d[t[i]]++;
  }
  int cnt = n;
  for (int i = 1; i <= n; i++) {
    if (d[i] < k) {
      q.push(i);
      vis[i] = 1;
      cnt--;
    }
  }
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (unsigned i = 0; i < edge[now].size(); i++) {
      int v = edge[now][i];
      if (S.count(mk(now, v))) continue;
      S.insert(mk(now, v));
      d[now]--, d[v]--;
      if (d[v] < k && vis[v] == 0) {
        q.push(v);
        vis[v] = 1;
        cnt--;
      }
    }
  }
  for (int i = m; i >= 1; --i) {
    ans[i] = cnt;
    if (vis[s[i]] || vis[t[i]] || S.count(mk(s[i], t[i]))) continue;
    S.insert(mk(s[i], t[i]));
    d[s[i]]--, d[t[i]]--;
    if (d[s[i]] < k && vis[s[i]] == 0) {
      q.push(s[i]);
      vis[s[i]] = 1;
      cnt--;
    }
    if (d[t[i]] < k && vis[t[i]] == 0) {
      q.push(t[i]);
      vis[t[i]] = 1;
      cnt--;
    }
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      for (unsigned i = 0; i < edge[now].size(); i++) {
        int v = edge[now][i];
        if (S.count(mk(now, v))) continue;
        d[now]--, d[v]--;
        if (d[v] < k && vis[v] == 0) {
          S.insert(mk(now, v));
          q.push(v);
          vis[v] = 1;
          cnt--;
        }
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```