#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n, m, k, l, i, ans, a[N], b[N], in[N], x[N], y[N], z[N], head[N], go[N + N],
    Next[N + N];
bool f[N], use[N + N];
inline void Add(int u, int v) {
  Next[++l] = head[u], head[u] = l, go[l] = v, use[l] = 1;
}
inline void bfs(int x) {
  int l, r, j, u, v;
  if (in[x] >= k || !f[x]) return;
  for (b[l = r = 1] = x, f[x] = 0, ans--; l <= r; l++) {
    for (j = head[u = b[l]]; j; j = Next[j]) {
      if (!use[j]) continue;
      in[v = go[j]]--, in[u]--;
      use[j] = use[j ^ 1] = 0;
      if (f[v] && in[v] < k) b[++r] = v, f[v] = 0, ans--;
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (l = 1, i = 1; i <= m; i++)
    scanf("%d%d", &x[i], &y[i]), Add(x[i], y[i]), Add(y[i], x[i]),
        z[i] = l, in[x[i]]++, in[y[i]]++;
  memset(f, 1, sizeof(f));
  for (ans = n, i = 1; i <= n; i++) bfs(i);
  for (i = m; i; i--) {
    a[i] = ans;
    if (!use[z[i]]) continue;
    in[x[i]]--, in[y[i]]--, use[z[i]] = use[z[i] ^ 1] = 0;
    bfs(x[i]), bfs(y[i]);
  }
  for (i = 1; i <= m; i++) printf("%d\n", a[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;
const int N = 200005;
int n, m, k, l, i, ans, a[N], b[N], in[N], x[N], y[N], z[N], head[N], go[N + N],
    Next[N + N];
bool f[N], use[N + N];
inline void Add(int u, int v) {
  Next[++l] = head[u], head[u] = l, go[l] = v, use[l] = 1;
}
inline void bfs(int x) {
  int l, r, j, u, v;
  if (in[x] >= k || !f[x]) return;
  for (b[l = r = 1] = x, f[x] = 0, ans--; l <= r; l++) {
    for (j = head[u = b[l]]; j; j = Next[j]) {
      if (!use[j]) continue;
      in[v = go[j]]--, in[u]--;
      use[j] = use[j ^ 1] = 0;
      if (f[v] && in[v] < k) b[++r] = v, f[v] = 0, ans--;
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (l = 1, i = 1; i <= m; i++)
    scanf("%d%d", &x[i], &y[i]), Add(x[i], y[i]), Add(y[i], x[i]),
        z[i] = l, in[x[i]]++, in[y[i]]++;
  memset(f, 1, sizeof(f));
  for (ans = n, i = 1; i <= n; i++) bfs(i);
  for (i = m; i; i--) {
    a[i] = ans;
    if (!use[z[i]]) continue;
    in[x[i]]--, in[y[i]]--, use[z[i]] = use[z[i] ^ 1] = 0;
    bfs(x[i]), bfs(y[i]);
  }
  for (i = 1; i <= m; i++) printf("%d\n", a[i]);
  return 0;
}

```