#include <bits/stdc++.h>
using namespace std;
int get() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-') {
    int s = 0;
    while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
    return -s;
  }
  int s = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
  return s;
}
const int N = 2e5 + 5;
struct edge {
  int x, nxt;
} e[N * 2];
int h[N], tot;
int d[N];
struct EDGE {
  int x, y;
} ed[N];
int n, m, k;
int ans[N];
int cnt;
bool bz[N];
bool er[N];
int update(int &now) {
  for (; now && er[now / 2]; now = e[now].nxt)
    ;
  return now;
}
void del(int x) {
  if (bz[x]) return;
  bz[x] = 1;
  cnt--;
  for (; update(h[x]);) {
    int p = h[x];
    d[x]--, d[e[p].x]--;
    er[p / 2] = 1;
    if (d[e[p].x] < k) del(e[p].x);
  }
}
void inse(int x, int y) {
  e[++tot].x = y;
  e[tot].nxt = h[x];
  h[x] = tot;
}
int main() {
  n = get();
  m = get();
  k = get();
  tot = 1;
  for (int i = 1; i <= m; i++) {
    ed[i].x = get();
    ed[i].y = get();
    inse(ed[i].x, ed[i].y);
    inse(ed[i].y, ed[i].x);
    d[ed[i].x]++, d[ed[i].y]++;
  }
  cnt = n;
  for (int i = 1; i <= n; i++)
    if (d[i] < k) del(i);
  for (int i = m; i >= 1; i--) {
    int x = ed[i].x, y = ed[i].y;
    ans[i] = cnt;
    if (er[i])
      ;
    else {
      er[i] = 1;
      d[x]--, d[y]--;
      if (d[x] < k) del(x);
      if (d[y] < k) del(y);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 get() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-') {
    int s = 0;
    while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
    return -s;
  }
  int s = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
  return s;
}
const int N = 2e5 + 5;
struct edge {
  int x, nxt;
} e[N * 2];
int h[N], tot;
int d[N];
struct EDGE {
  int x, y;
} ed[N];
int n, m, k;
int ans[N];
int cnt;
bool bz[N];
bool er[N];
int update(int &now) {
  for (; now && er[now / 2]; now = e[now].nxt)
    ;
  return now;
}
void del(int x) {
  if (bz[x]) return;
  bz[x] = 1;
  cnt--;
  for (; update(h[x]);) {
    int p = h[x];
    d[x]--, d[e[p].x]--;
    er[p / 2] = 1;
    if (d[e[p].x] < k) del(e[p].x);
  }
}
void inse(int x, int y) {
  e[++tot].x = y;
  e[tot].nxt = h[x];
  h[x] = tot;
}
int main() {
  n = get();
  m = get();
  k = get();
  tot = 1;
  for (int i = 1; i <= m; i++) {
    ed[i].x = get();
    ed[i].y = get();
    inse(ed[i].x, ed[i].y);
    inse(ed[i].y, ed[i].x);
    d[ed[i].x]++, d[ed[i].y]++;
  }
  cnt = n;
  for (int i = 1; i <= n; i++)
    if (d[i] < k) del(i);
  for (int i = m; i >= 1; i--) {
    int x = ed[i].x, y = ed[i].y;
    ans[i] = cnt;
    if (er[i])
      ;
    else {
      er[i] = 1;
      d[x]--, d[y]--;
      if (d[x] < k) del(x);
      if (d[y] < k) del(y);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```