#include <bits/stdc++.h>
using namespace std;
int n, m, k;
struct edge {
  int v, nxt;
} e[200005 * 2];
int cnt = 1, head[200005];
void adde(int u, int v) {
  e[++cnt].v = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
int vis[200005 * 2], d[200005], ans[200005], res;
struct ege2 {
  int u, v;
} edge[200005];
queue<int> q;
void solve() {
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    d[u] = 0;
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].v;
      if (vis[i]) continue;
      vis[i] = vis[i ^ 1] = 1;
      d[v]--;
      if (d[v] == k - 1) {
        --res;
        q.push(v);
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, u, v; i <= m; ++i) {
    scanf("%d%d", &u, &v);
    edge[i].u = u;
    edge[i].v = v;
    adde(u, v);
    adde(v, u);
    ++d[u];
    ++d[v];
  }
  res = n;
  for (int i = 1; i <= n; ++i)
    if (d[i] < k) {
      --res;
      q.push(i);
    }
  solve();
  ans[m] = res;
  for (int i = m; i >= 2; --i) {
    if (vis[i << 1]) {
      ans[i - 1] = ans[i];
      continue;
    }
    vis[i << 1] = vis[i << 1 | 1] = 1;
    int u = edge[i].u, v = edge[i].v;
    d[u]--;
    d[v]--;
    if (d[u] == k - 1) {
      --res;
      q.push(u);
    }
    if (d[v] == k - 1) {
      --res;
      q.push(v);
    }
    solve();
    ans[i - 1] = res;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Develop a solution in cpp 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;
int n, m, k;
struct edge {
  int v, nxt;
} e[200005 * 2];
int cnt = 1, head[200005];
void adde(int u, int v) {
  e[++cnt].v = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
int vis[200005 * 2], d[200005], ans[200005], res;
struct ege2 {
  int u, v;
} edge[200005];
queue<int> q;
void solve() {
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    d[u] = 0;
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].v;
      if (vis[i]) continue;
      vis[i] = vis[i ^ 1] = 1;
      d[v]--;
      if (d[v] == k - 1) {
        --res;
        q.push(v);
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, u, v; i <= m; ++i) {
    scanf("%d%d", &u, &v);
    edge[i].u = u;
    edge[i].v = v;
    adde(u, v);
    adde(v, u);
    ++d[u];
    ++d[v];
  }
  res = n;
  for (int i = 1; i <= n; ++i)
    if (d[i] < k) {
      --res;
      q.push(i);
    }
  solve();
  ans[m] = res;
  for (int i = m; i >= 2; --i) {
    if (vis[i << 1]) {
      ans[i - 1] = ans[i];
      continue;
    }
    vis[i << 1] = vis[i << 1 | 1] = 1;
    int u = edge[i].u, v = edge[i].v;
    d[u]--;
    d[v]--;
    if (d[u] == k - 1) {
      --res;
      q.push(u);
    }
    if (d[v] == k - 1) {
      --res;
      q.push(v);
    }
    solve();
    ans[i - 1] = res;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

```