#include <bits/stdc++.h>
using namespace std;
struct Edge {
  int x, y;
} E[200010];
struct edge {
  int nxt, t, s;
} e[200010 << 1];
int head[200010], edge_cnt;
void add_edge(int x, int y, int z) {
  e[edge_cnt] = (edge){head[x], y, z};
  head[x] = edge_cnt++;
}
bool ban[200010];
int Deg[200010], Q[200010], Ans[200010];
int main() {
  memset(head, -1, sizeof(head));
  int n, m, K, i, j;
  scanf("%d%d%d", &n, &m, &K);
  for (i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y, i);
    add_edge(y, x, i);
    E[i] = (Edge){x, y};
    Deg[x]++, Deg[y]++;
  }
  int L = 1, R = 0;
  for (i = 1; i <= n; i++) {
    if (Deg[i] < K) {
      Q[++R] = i;
      ban[i] = 1;
    }
  }
  for (i = m; i >= 1; i--) {
    while (L <= R) {
      int x = Q[L++];
      for (j = head[x]; ~j; j = e[j].nxt) {
        int y = e[j].t;
        if (ban[y] || e[j].s > i) {
          continue;
        }
        Deg[y]--;
        if (Deg[y] < K) {
          Q[++R] = y;
          ban[y] = 1;
        }
      }
    }
    Ans[i] = n - R;
    int a = E[i].x, b = E[i].y;
    if (ban[a] || ban[b]) {
      continue;
    }
    Deg[a]--, Deg[b]--;
    if (Deg[a] < K) {
      Q[++R] = a;
      ban[a] = 1;
    }
    if (Deg[b] < K) {
      Q[++R] = b;
      ban[b] = 1;
    }
  }
  for (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>
using namespace std;
struct Edge {
  int x, y;
} E[200010];
struct edge {
  int nxt, t, s;
} e[200010 << 1];
int head[200010], edge_cnt;
void add_edge(int x, int y, int z) {
  e[edge_cnt] = (edge){head[x], y, z};
  head[x] = edge_cnt++;
}
bool ban[200010];
int Deg[200010], Q[200010], Ans[200010];
int main() {
  memset(head, -1, sizeof(head));
  int n, m, K, i, j;
  scanf("%d%d%d", &n, &m, &K);
  for (i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y, i);
    add_edge(y, x, i);
    E[i] = (Edge){x, y};
    Deg[x]++, Deg[y]++;
  }
  int L = 1, R = 0;
  for (i = 1; i <= n; i++) {
    if (Deg[i] < K) {
      Q[++R] = i;
      ban[i] = 1;
    }
  }
  for (i = m; i >= 1; i--) {
    while (L <= R) {
      int x = Q[L++];
      for (j = head[x]; ~j; j = e[j].nxt) {
        int y = e[j].t;
        if (ban[y] || e[j].s > i) {
          continue;
        }
        Deg[y]--;
        if (Deg[y] < K) {
          Q[++R] = y;
          ban[y] = 1;
        }
      }
    }
    Ans[i] = n - R;
    int a = E[i].x, b = E[i].y;
    if (ban[a] || ban[b]) {
      continue;
    }
    Deg[a]--, Deg[b]--;
    if (Deg[a] < K) {
      Q[++R] = a;
      ban[a] = 1;
    }
    if (Deg[b] < K) {
      Q[++R] = b;
      ban[b] = 1;
    }
  }
  for (i = 1; i <= m; i++) {
    printf("%d\n", Ans[i]);
  }
  return 0;
}

```