#include <bits/stdc++.h>
using namespace std;
struct Edge {
  int t, id, next;
  Edge() {}
  Edge(int a, int b, int c) : t(a), id(b), next(c) {}
};
Edge e[400005];
int head[200005], ans[200005], k;
int cur;
int d[200005], sum;
bool col[200005];
void dfs(int x) {
  col[x] = 0;
  sum--;
  for (int i = head[x]; i; i = e[i].next)
    if (col[e[i].t] && e[i].id < cur) {
      int u = e[i].t;
      d[u]--;
      if (d[u] < k) dfs(u);
    }
}
pair<int, int> a[200005];
int main() {
  int n, m;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[2 * i - 1] = Edge(y, i, head[x]);
    head[x] = 2 * i - 1;
    e[2 * i] = Edge(x, i, head[y]);
    head[y] = 2 * i;
    d[x]++;
    d[y]++;
    a[i] = pair<int, int>(x, y);
  }
  for (int i = 1; i <= n; i++) col[i] = 1;
  sum = n;
  cur = m + 1;
  for (int i = 1; i <= n; i++)
    if (col[i] && d[i] < k) dfs(i);
  for (int i = m; i > 0; i--) {
    ans[i] = sum;
    int x = a[i].first, y = a[i].second;
    cur--;
    if (col[x] && col[y]) {
      d[x]--;
      d[y]--;
      if (col[x] && d[x] < k) dfs(x);
      if (col[y] && d[y] < k) dfs(y);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### 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 t, id, next;
  Edge() {}
  Edge(int a, int b, int c) : t(a), id(b), next(c) {}
};
Edge e[400005];
int head[200005], ans[200005], k;
int cur;
int d[200005], sum;
bool col[200005];
void dfs(int x) {
  col[x] = 0;
  sum--;
  for (int i = head[x]; i; i = e[i].next)
    if (col[e[i].t] && e[i].id < cur) {
      int u = e[i].t;
      d[u]--;
      if (d[u] < k) dfs(u);
    }
}
pair<int, int> a[200005];
int main() {
  int n, m;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[2 * i - 1] = Edge(y, i, head[x]);
    head[x] = 2 * i - 1;
    e[2 * i] = Edge(x, i, head[y]);
    head[y] = 2 * i;
    d[x]++;
    d[y]++;
    a[i] = pair<int, int>(x, y);
  }
  for (int i = 1; i <= n; i++) col[i] = 1;
  sum = n;
  cur = m + 1;
  for (int i = 1; i <= n; i++)
    if (col[i] && d[i] < k) dfs(i);
  for (int i = m; i > 0; i--) {
    ans[i] = sum;
    int x = a[i].first, y = a[i].second;
    cur--;
    if (col[x] && col[y]) {
      d[x]--;
      d[y]--;
      if (col[x] && d[x] < k) dfs(x);
      if (col[y] && d[y] < k) dfs(y);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```