#include <bits/stdc++.h>
using std::queue;
const int N = 2e5 + 6;
template <class T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
}
struct Edge {
  int u, v, id;
} e[N << 1];
int head[N], ecnt;
inline void addedge(int u, int v, int id) {
  e[++ecnt].v = v;
  e[ecnt].u = head[u];
  head[u] = ecnt;
  e[ecnt].id = id;
}
inline void add(int u, int v, int id) {
  addedge(u, v, id);
  addedge(v, u, id);
}
int n, m, k, du[N], a[N], b[N], del[N], u, ans, inq[N], Ans[N], len;
queue<int> q;
signed main() {
  read(n);
  read(m);
  read(k);
  for (int i = 1; i <= m; i++) {
    read(a[i]);
    read(b[i]);
    du[a[i]]++;
    du[b[i]]++;
    add(a[i], b[i], i);
  }
  for (int i = 1; i <= n; i++)
    if (du[i] < k) q.push(i), inq[i] = 1;
  ans = n;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    ans--;
    du[u] = 0;
    for (int i = head[u], v; i && (v = e[i].v); i = e[i].u) {
      if (inq[v]) continue;
      if (del[e[i].id]) continue;
      du[v]--;
      del[e[i].id] = 1;
      if (du[v] < k) q.push(v), inq[v] = 1;
    }
  }
  for (int i = m; i >= 1; i--) {
    Ans[++len] = ans;
    if (del[i]) continue;
    du[a[i]]--;
    du[b[i]]--;
    del[i] = 1;
    if (!inq[a[i]] && du[a[i]] < k) q.push(a[i]), inq[a[i]] = 1;
    if (!inq[b[i]] && du[b[i]] < k) q.push(b[i]), inq[b[i]] = 1;
    while (!q.empty()) {
      u = q.front();
      q.pop();
      ans--;
      du[u] = 0;
      for (int i = head[u], v; i && (v = e[i].v); i = e[i].u) {
        if (inq[v]) continue;
        if (del[e[i].id]) continue;
        du[v]--;
        del[e[i].id] = 1;
        if (du[v] < k) q.push(v), inq[v] = 1;
      }
    }
  }
  for (int i = m; i >= 1; i--) printf("%d\n", Ans[i]);
}

### Prompt

Please create a solution in Cpp 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 std::queue;
const int N = 2e5 + 6;
template <class T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
}
struct Edge {
  int u, v, id;
} e[N << 1];
int head[N], ecnt;
inline void addedge(int u, int v, int id) {
  e[++ecnt].v = v;
  e[ecnt].u = head[u];
  head[u] = ecnt;
  e[ecnt].id = id;
}
inline void add(int u, int v, int id) {
  addedge(u, v, id);
  addedge(v, u, id);
}
int n, m, k, du[N], a[N], b[N], del[N], u, ans, inq[N], Ans[N], len;
queue<int> q;
signed main() {
  read(n);
  read(m);
  read(k);
  for (int i = 1; i <= m; i++) {
    read(a[i]);
    read(b[i]);
    du[a[i]]++;
    du[b[i]]++;
    add(a[i], b[i], i);
  }
  for (int i = 1; i <= n; i++)
    if (du[i] < k) q.push(i), inq[i] = 1;
  ans = n;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    ans--;
    du[u] = 0;
    for (int i = head[u], v; i && (v = e[i].v); i = e[i].u) {
      if (inq[v]) continue;
      if (del[e[i].id]) continue;
      du[v]--;
      del[e[i].id] = 1;
      if (du[v] < k) q.push(v), inq[v] = 1;
    }
  }
  for (int i = m; i >= 1; i--) {
    Ans[++len] = ans;
    if (del[i]) continue;
    du[a[i]]--;
    du[b[i]]--;
    del[i] = 1;
    if (!inq[a[i]] && du[a[i]] < k) q.push(a[i]), inq[a[i]] = 1;
    if (!inq[b[i]] && du[b[i]] < k) q.push(b[i]), inq[b[i]] = 1;
    while (!q.empty()) {
      u = q.front();
      q.pop();
      ans--;
      du[u] = 0;
      for (int i = head[u], v; i && (v = e[i].v); i = e[i].u) {
        if (inq[v]) continue;
        if (del[e[i].id]) continue;
        du[v]--;
        del[e[i].id] = 1;
        if (du[v] < k) q.push(v), inq[v] = 1;
      }
    }
  }
  for (int i = m; i >= 1; i--) printf("%d\n", Ans[i]);
}

```