#include <bits/stdc++.h>
using namespace std;
struct Edge {
  int u, v, nxt;
} e[400010];
int tot;
int first[200010];
int d[200010];
void build(int u, int v) {
  e[++tot] = (Edge){u, v, first[u]};
  first[u] = tot;
  d[u]++;
  return;
}
int ans[200010];
bool book[200010];
bool is[200010];
int n, m, k;
queue<int> q;
void getans(int w) {
  if (w < m) {
    ans[w] = ans[w + 1];
    int u = e[(w + 1) * 2].u, v = e[(w + 1) * 2].v;
    if (book[w + 1]) return;
    d[u]--;
    d[v]--;
    book[w + 1] = true;
    if (!is[u] && d[u] < k) q.push(u), ans[w]--, is[u] = true;
    if (!is[v] && d[v] < k) q.push(v), ans[w]--, is[v] = true;
  } else {
    ans[w] = n;
    for (int i = 1; i <= n; i++)
      if (d[i] < k) q.push(i), ans[w]--, is[i] = true;
  }
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = first[now]; i; i = e[i].nxt) {
      if (book[(i + 1) / 2]) continue;
      if (is[e[i].v]) continue;
      book[(i + 1) / 2] = true;
      d[e[i].v]--;
      if (d[e[i].v] < k) {
        q.push(e[i].v);
        ans[w]--;
        is[e[i].v] = true;
      }
    }
  }
  return;
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    build(u, v);
    build(v, u);
  }
  for (int i = m; i >= 1; i--) getans(i);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
struct Edge {
  int u, v, nxt;
} e[400010];
int tot;
int first[200010];
int d[200010];
void build(int u, int v) {
  e[++tot] = (Edge){u, v, first[u]};
  first[u] = tot;
  d[u]++;
  return;
}
int ans[200010];
bool book[200010];
bool is[200010];
int n, m, k;
queue<int> q;
void getans(int w) {
  if (w < m) {
    ans[w] = ans[w + 1];
    int u = e[(w + 1) * 2].u, v = e[(w + 1) * 2].v;
    if (book[w + 1]) return;
    d[u]--;
    d[v]--;
    book[w + 1] = true;
    if (!is[u] && d[u] < k) q.push(u), ans[w]--, is[u] = true;
    if (!is[v] && d[v] < k) q.push(v), ans[w]--, is[v] = true;
  } else {
    ans[w] = n;
    for (int i = 1; i <= n; i++)
      if (d[i] < k) q.push(i), ans[w]--, is[i] = true;
  }
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = first[now]; i; i = e[i].nxt) {
      if (book[(i + 1) / 2]) continue;
      if (is[e[i].v]) continue;
      book[(i + 1) / 2] = true;
      d[e[i].v]--;
      if (d[e[i].v] < k) {
        q.push(e[i].v);
        ans[w]--;
        is[e[i].v] = true;
      }
    }
  }
  return;
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    build(u, v);
    build(v, u);
  }
  for (int i = m; i >= 1; i--) getans(i);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```