#include <bits/stdc++.h>
using namespace std;
int n, m, cnt, k, tot, tp;
int head[200005];
int sta[200005];
int dgr[200005];
int ans[200005];
bool usd[200005];
struct node {
  int fr;
  int to;
  int nxt;
  int mrk;
} edge[400005];
void init() {
  cnt = 0;
  memset(head, -1, sizeof(head));
}
void addedge(int f, int t) {
  cnt++;
  edge[cnt].fr = f;
  edge[cnt].to = t;
  edge[cnt].nxt = head[f];
  head[f] = cnt;
}
queue<int> que;
int main() {
  init();
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
    dgr[v]++;
    dgr[u]++;
  }
  for (int i = 1; i <= n; i++) {
    if (dgr[i] < k) {
      que.push(i);
    }
  }
  tot = n;
  tp = m;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (usd[u]) continue;
    usd[u] = true;
    tot--;
    for (int i = head[u]; i != -1; i = edge[i].nxt) {
      int v = edge[i].to;
      dgr[v]--;
      if (dgr[v] < k) {
        que.push(v);
      }
    }
  }
  for (int i = cnt - 1; i >= 1; i -= 2) {
    ans[tp--] = tot;
    int u = edge[i].fr;
    int v = edge[i].to;
    if (usd[u] || usd[v]) continue;
    dgr[u]--, dgr[v]--;
    edge[i].mrk = edge[i + 1].mrk = 1;
    if (dgr[u] < k) que.push(u);
    if (dgr[v] < k) que.push(v);
    while (!que.empty()) {
      int s = que.front();
      que.pop();
      if (usd[s]) continue;
      usd[s] = true;
      tot--;
      for (int i = head[s]; i != -1; i = edge[i].nxt) {
        if (edge[i].mrk) continue;
        int v = edge[i].to;
        dgr[v]--;
        if (dgr[v] < k) {
          que.push(v);
        }
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution 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;
int n, m, cnt, k, tot, tp;
int head[200005];
int sta[200005];
int dgr[200005];
int ans[200005];
bool usd[200005];
struct node {
  int fr;
  int to;
  int nxt;
  int mrk;
} edge[400005];
void init() {
  cnt = 0;
  memset(head, -1, sizeof(head));
}
void addedge(int f, int t) {
  cnt++;
  edge[cnt].fr = f;
  edge[cnt].to = t;
  edge[cnt].nxt = head[f];
  head[f] = cnt;
}
queue<int> que;
int main() {
  init();
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
    dgr[v]++;
    dgr[u]++;
  }
  for (int i = 1; i <= n; i++) {
    if (dgr[i] < k) {
      que.push(i);
    }
  }
  tot = n;
  tp = m;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (usd[u]) continue;
    usd[u] = true;
    tot--;
    for (int i = head[u]; i != -1; i = edge[i].nxt) {
      int v = edge[i].to;
      dgr[v]--;
      if (dgr[v] < k) {
        que.push(v);
      }
    }
  }
  for (int i = cnt - 1; i >= 1; i -= 2) {
    ans[tp--] = tot;
    int u = edge[i].fr;
    int v = edge[i].to;
    if (usd[u] || usd[v]) continue;
    dgr[u]--, dgr[v]--;
    edge[i].mrk = edge[i + 1].mrk = 1;
    if (dgr[u] < k) que.push(u);
    if (dgr[v] < k) que.push(v);
    while (!que.empty()) {
      int s = que.front();
      que.pop();
      if (usd[s]) continue;
      usd[s] = true;
      tot--;
      for (int i = head[s]; i != -1; i = edge[i].nxt) {
        if (edge[i].mrk) continue;
        int v = edge[i].to;
        dgr[v]--;
        if (dgr[v] < k) {
          que.push(v);
        }
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```