#include <bits/stdc++.h>
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c <= '9' && c >= '0') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
void print(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 5000007;
int n, m, k;
int d[maxn];
bool del[maxn];
using namespace std;
map<pair<int, int>, int> mp;
struct node {
  int v, next;
} edge[maxn];
int head[maxn], num = 0;
inline void add_edge(int u, int v) {
  edge[++num].v = v;
  edge[num].next = head[u];
  head[u] = num;
}
int ans;
int u[maxn], v[maxn];
int Ans[maxn];
queue<int> q;
void solve(int x) {
  if (d[x] >= k || del[x]) return;
  del[x] = 1;
  q.push(x);
  ans--;
  while (!q.empty()) {
    int U = q.front();
    q.pop();
    for (int i = head[U]; i; i = edge[i].next) {
      int V = edge[i].v;
      if (del[V]) continue;
      if (mp.count(make_pair(U, V)) == 0) d[V]--;
      if (d[V] < k) {
        del[V] = true;
        ans--;
        q.push(V);
      }
    }
  }
}
int main() {
  n = read(), m = read();
  k = read();
  for (int i = 1; i <= m; ++i) {
    u[i] = read(), v[i] = read();
    add_edge(u[i], v[i]);
    add_edge(v[i], u[i]);
    d[v[i]]++;
    d[u[i]]++;
  }
  ans = n;
  for (int i = 1; i <= n; ++i) solve(i);
  mp.clear();
  for (int i = m; i >= 1; --i) {
    Ans[i] = ans;
    if (!del[u[i]]) d[v[i]]--;
    if (!del[v[i]]) d[u[i]]--;
    mp[pair<int, int>(u[i], v[i])] = 1;
    mp[pair<int, int>(v[i], u[i])] = 1;
    solve(u[i]);
    solve(v[i]);
  }
  for (int i = 1; i <= m; ++i) print(Ans[i]), putchar('\n');
}

### Prompt

Please formulate a CPP solution 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>
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c <= '9' && c >= '0') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
void print(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 5000007;
int n, m, k;
int d[maxn];
bool del[maxn];
using namespace std;
map<pair<int, int>, int> mp;
struct node {
  int v, next;
} edge[maxn];
int head[maxn], num = 0;
inline void add_edge(int u, int v) {
  edge[++num].v = v;
  edge[num].next = head[u];
  head[u] = num;
}
int ans;
int u[maxn], v[maxn];
int Ans[maxn];
queue<int> q;
void solve(int x) {
  if (d[x] >= k || del[x]) return;
  del[x] = 1;
  q.push(x);
  ans--;
  while (!q.empty()) {
    int U = q.front();
    q.pop();
    for (int i = head[U]; i; i = edge[i].next) {
      int V = edge[i].v;
      if (del[V]) continue;
      if (mp.count(make_pair(U, V)) == 0) d[V]--;
      if (d[V] < k) {
        del[V] = true;
        ans--;
        q.push(V);
      }
    }
  }
}
int main() {
  n = read(), m = read();
  k = read();
  for (int i = 1; i <= m; ++i) {
    u[i] = read(), v[i] = read();
    add_edge(u[i], v[i]);
    add_edge(v[i], u[i]);
    d[v[i]]++;
    d[u[i]]++;
  }
  ans = n;
  for (int i = 1; i <= n; ++i) solve(i);
  mp.clear();
  for (int i = m; i >= 1; --i) {
    Ans[i] = ans;
    if (!del[u[i]]) d[v[i]]--;
    if (!del[v[i]]) d[u[i]]--;
    mp[pair<int, int>(u[i], v[i])] = 1;
    mp[pair<int, int>(v[i], u[i])] = 1;
    solve(u[i]);
    solve(v[i]);
  }
  for (int i = 1; i <= m; ++i) print(Ans[i]), putchar('\n');
}

```