#include <bits/stdc++.h>
using namespace std;
struct edge {
  int from, to, vis, next;
} e[400020];
int top, head[200020];
vector<int> g[200020];
int in[200020], gg[200020];
int n, m, k;
int ans[200020], cnt;
void add_edge(int from, int to) {
  e[++top].to = to;
  e[top].from = from;
  e[top].next = head[from];
  head[from] = top;
}
void add(int from, int to) {
  add_edge(from, to);
  add_edge(to, from);
}
queue<int> q;
void del(int pos) {
  if (!gg[pos]) cnt--;
  q.push(pos);
  gg[pos] = 1;
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = head[now]; ~i; i = e[i].next) {
      if (e[i].vis || gg[e[i].to]) continue;
      in[e[i].to]--;
      e[i].vis = 1;
      e[i ^ 1].vis = 1;
      if (in[e[i].to] < k) {
        cnt--;
        gg[e[i].to] = 1;
        q.push(e[i].to);
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  cnt = n;
  top = -1;
  memset(head, -1, sizeof(head));
  int from, to;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &from, &to);
    in[from]++, in[to]++;
    add(from, to);
  }
  for (int i = 1; i <= n; i++) {
    if (head[i] == -1) cnt--, gg[i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (in[i] < k && !gg[i]) del(i);
  }
  int ttt = m;
  for (int i = 2 * m - 1; i >= 1; i -= 2) {
    ans[ttt--] = cnt;
    if (e[i].vis) continue;
    in[e[i].from]--;
    in[e[i].to]--;
    e[i].vis = e[i ^ 1].vis = 1;
    if (!gg[e[i].from] && in[e[i].from] < k) del(e[i].from);
    if (!gg[e[i].to] && in[e[i].to] < k) del(e[i].to);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}

### Prompt

Please provide a CPP coded solution 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;
struct edge {
  int from, to, vis, next;
} e[400020];
int top, head[200020];
vector<int> g[200020];
int in[200020], gg[200020];
int n, m, k;
int ans[200020], cnt;
void add_edge(int from, int to) {
  e[++top].to = to;
  e[top].from = from;
  e[top].next = head[from];
  head[from] = top;
}
void add(int from, int to) {
  add_edge(from, to);
  add_edge(to, from);
}
queue<int> q;
void del(int pos) {
  if (!gg[pos]) cnt--;
  q.push(pos);
  gg[pos] = 1;
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = head[now]; ~i; i = e[i].next) {
      if (e[i].vis || gg[e[i].to]) continue;
      in[e[i].to]--;
      e[i].vis = 1;
      e[i ^ 1].vis = 1;
      if (in[e[i].to] < k) {
        cnt--;
        gg[e[i].to] = 1;
        q.push(e[i].to);
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  cnt = n;
  top = -1;
  memset(head, -1, sizeof(head));
  int from, to;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &from, &to);
    in[from]++, in[to]++;
    add(from, to);
  }
  for (int i = 1; i <= n; i++) {
    if (head[i] == -1) cnt--, gg[i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (in[i] < k && !gg[i]) del(i);
  }
  int ttt = m;
  for (int i = 2 * m - 1; i >= 1; i -= 2) {
    ans[ttt--] = cnt;
    if (e[i].vis) continue;
    in[e[i].from]--;
    in[e[i].to]--;
    e[i].vis = e[i ^ 1].vis = 1;
    if (!gg[e[i].from] && in[e[i].from] < k) del(e[i].from);
    if (!gg[e[i].to] && in[e[i].to] < k) del(e[i].to);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}

```