#include <bits/stdc++.h>
struct pnt {
  int hd;
  int ind;
  bool ded;
} p[1000000];
struct ent {
  int twd;
  int lst;
} e[1000000];
int cnt;
int n, m, k;
int lft;
int ans[1000000];
int u[1000000], v[1000000];
void ade(int f, int t) {
  cnt++;
  e[cnt].twd = t;
  e[cnt].lst = p[f].hd;
  p[f].hd = cnt;
  p[f].ind++;
  return;
}
void Delete(int x) {
  if (p[x].ded) return;
  p[x].ded = true;
  lft--;
  for (int i = p[x].hd; i; i = e[i].lst) {
    int to = e[i].twd;
    p[to].ind--;
    if (p[to].ind < k) Delete(to);
  }
  return;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u[i], &v[i]);
    ade(u[i], v[i]);
    ade(v[i], u[i]);
  }
  lft = n;
  for (int i = 1; i <= n; i++) {
    if (p[i].ind < k) {
      Delete(i);
    }
  }
  for (int i = m; i; i--) {
    ans[i] = lft;
    p[u[i]].hd = e[p[u[i]].hd].lst;
    p[v[i]].hd = e[p[v[i]].hd].lst;
    if (!p[v[i]].ded) p[u[i]].ind--;
    if (!p[u[i]].ded) p[v[i]].ind--;
    if (p[v[i]].ind < k) Delete(v[i]);
    if (p[u[i]].ind < k) Delete(u[i]);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please create a solution in cpp 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>
struct pnt {
  int hd;
  int ind;
  bool ded;
} p[1000000];
struct ent {
  int twd;
  int lst;
} e[1000000];
int cnt;
int n, m, k;
int lft;
int ans[1000000];
int u[1000000], v[1000000];
void ade(int f, int t) {
  cnt++;
  e[cnt].twd = t;
  e[cnt].lst = p[f].hd;
  p[f].hd = cnt;
  p[f].ind++;
  return;
}
void Delete(int x) {
  if (p[x].ded) return;
  p[x].ded = true;
  lft--;
  for (int i = p[x].hd; i; i = e[i].lst) {
    int to = e[i].twd;
    p[to].ind--;
    if (p[to].ind < k) Delete(to);
  }
  return;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u[i], &v[i]);
    ade(u[i], v[i]);
    ade(v[i], u[i]);
  }
  lft = n;
  for (int i = 1; i <= n; i++) {
    if (p[i].ind < k) {
      Delete(i);
    }
  }
  for (int i = m; i; i--) {
    ans[i] = lft;
    p[u[i]].hd = e[p[u[i]].hd].lst;
    p[v[i]].hd = e[p[v[i]].hd].lst;
    if (!p[v[i]].ded) p[u[i]].ind--;
    if (!p[u[i]].ded) p[v[i]].ind--;
    if (p[v[i]].ind < k) Delete(v[i]);
    if (p[u[i]].ind < k) Delete(u[i]);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```