#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 5;
struct edge {
  int u, v, ne, num;
} e[maxn];
int n, m, k, sum;
int d[maxn], no[maxn], vis[maxn];
int head[maxn], len;
int ans[maxn];
void add(int u, int v, int num) {
  e[len].u = u;
  e[len].v = v;
  e[len].num = num;
  e[len].ne = head[u];
  head[u] = len++;
}
void del(int x) {
  sum--;
  no[x] = 1;
  for (int i = head[x]; i != -1; i = e[i].ne) {
    int v = e[i].v;
    if (no[v] || vis[e[i].num]) continue;
    vis[e[i].num] = 1;
    d[v]--;
    if (d[v] < k) del(v);
  }
  return;
}
int main() {
  memset(head, -1, sizeof(head));
  cin >> n >> m >> k;
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d %d", &x, &y);
    add(x, y, i);
    add(y, x, i);
    d[x]++;
    d[y]++;
  }
  sum = n;
  for (int i = 1; i <= n; i++) {
    if (d[i] >= k || no[i]) continue;
    del(i);
  }
  ans[m] = sum;
  for (int i = len - 1; i > 0; i -= 2) {
    int u = e[i].u, v = e[i].v;
    if (!vis[e[i].num]) {
      vis[e[i].num] = 1;
      if (!no[u]) {
        d[u]--;
        if (d[u] < k) del(u);
      }
      if (!no[v]) {
        d[v]--;
        if (d[v] < k) del(v);
      }
    }
    ans[(i + 1) / 2 - 1] = sum;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 5;
struct edge {
  int u, v, ne, num;
} e[maxn];
int n, m, k, sum;
int d[maxn], no[maxn], vis[maxn];
int head[maxn], len;
int ans[maxn];
void add(int u, int v, int num) {
  e[len].u = u;
  e[len].v = v;
  e[len].num = num;
  e[len].ne = head[u];
  head[u] = len++;
}
void del(int x) {
  sum--;
  no[x] = 1;
  for (int i = head[x]; i != -1; i = e[i].ne) {
    int v = e[i].v;
    if (no[v] || vis[e[i].num]) continue;
    vis[e[i].num] = 1;
    d[v]--;
    if (d[v] < k) del(v);
  }
  return;
}
int main() {
  memset(head, -1, sizeof(head));
  cin >> n >> m >> k;
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d %d", &x, &y);
    add(x, y, i);
    add(y, x, i);
    d[x]++;
    d[y]++;
  }
  sum = n;
  for (int i = 1; i <= n; i++) {
    if (d[i] >= k || no[i]) continue;
    del(i);
  }
  ans[m] = sum;
  for (int i = len - 1; i > 0; i -= 2) {
    int u = e[i].u, v = e[i].v;
    if (!vis[e[i].num]) {
      vis[e[i].num] = 1;
      if (!no[u]) {
        d[u]--;
        if (d[u] < k) del(u);
      }
      if (!no[v]) {
        d[v]--;
        if (d[v] < k) del(v);
      }
    }
    ans[(i + 1) / 2 - 1] = sum;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```