#include <bits/stdc++.h>
using namespace std;
const long double Pi = 3.14159265359;
const long long MOD = (long long)1e9 + 7;
const long long MAXN = (long long)2e5 + 10;
const long long INF = (long long)9223372036854775;
const long double EPS = (long double)1e-8;
set<long long> ans;
set<long long> g[MAXN];
vector<pair<long long, long long> > q;
vector<long long> anss;
queue<long long> bfsq;
long long vis[MAXN];
int main() {
  long long n, m, k, u, v;
  scanf("%lld", &n);
  scanf("%lld", &m);
  scanf("%lld", &k);
  long long an = n;
  for (int i = 0; i < m; i++) {
    scanf("%lld", &u);
    scanf("%lld", &v);
    u--;
    v--;
    g[u].insert(v);
    g[v].insert(u);
    q.push_back(make_pair(u, v));
  }
  for (int i = 0; i < n; i++) ans.insert(i);
  for (int i = 0; i < n; i++) {
    if (g[i].size() < k) {
      bfsq.push(i);
      vis[i] = 1;
    }
  }
  while (bfsq.size() != 0) {
    for (auto adj : g[bfsq.front()]) {
      g[adj].erase(bfsq.front());
      if (g[adj].size() < k) {
        if (vis[adj] == 0) bfsq.push(adj);
        vis[adj] = 1;
      }
    }
    ans.erase(bfsq.front());
    bfsq.pop();
    an--;
  }
  anss.push_back(an);
  for (int i = 0; i < m; i++) {
    u = q.back().first;
    v = q.back().second;
    if ((vis[u] == 1) or (vis[v] == 1)) {
      anss.push_back(an);
      q.pop_back();
      continue;
    }
    g[u].erase(v);
    g[v].erase(u);
    if (g[u].size() < k) {
      bfsq.push(u);
      vis[u] = 1;
    }
    if (g[v].size() < k) {
      bfsq.push(v);
      vis[v] = 1;
    }
    while (bfsq.size() != 0) {
      for (auto adj : g[bfsq.front()]) {
        g[adj].erase(bfsq.front());
        if (g[adj].size() < k) {
          if (vis[adj] == 0) bfsq.push(adj);
          vis[adj] = 1;
        }
      }
      ans.erase(bfsq.front());
      bfsq.pop();
      an--;
    }
    anss.push_back(an);
    q.pop_back();
  }
  anss.pop_back();
  for (int i = 0; i < m; i++) {
    printf("%lld\n", anss.back());
    anss.pop_back();
  }
  return 0;
}

### Prompt

Generate 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>
using namespace std;
const long double Pi = 3.14159265359;
const long long MOD = (long long)1e9 + 7;
const long long MAXN = (long long)2e5 + 10;
const long long INF = (long long)9223372036854775;
const long double EPS = (long double)1e-8;
set<long long> ans;
set<long long> g[MAXN];
vector<pair<long long, long long> > q;
vector<long long> anss;
queue<long long> bfsq;
long long vis[MAXN];
int main() {
  long long n, m, k, u, v;
  scanf("%lld", &n);
  scanf("%lld", &m);
  scanf("%lld", &k);
  long long an = n;
  for (int i = 0; i < m; i++) {
    scanf("%lld", &u);
    scanf("%lld", &v);
    u--;
    v--;
    g[u].insert(v);
    g[v].insert(u);
    q.push_back(make_pair(u, v));
  }
  for (int i = 0; i < n; i++) ans.insert(i);
  for (int i = 0; i < n; i++) {
    if (g[i].size() < k) {
      bfsq.push(i);
      vis[i] = 1;
    }
  }
  while (bfsq.size() != 0) {
    for (auto adj : g[bfsq.front()]) {
      g[adj].erase(bfsq.front());
      if (g[adj].size() < k) {
        if (vis[adj] == 0) bfsq.push(adj);
        vis[adj] = 1;
      }
    }
    ans.erase(bfsq.front());
    bfsq.pop();
    an--;
  }
  anss.push_back(an);
  for (int i = 0; i < m; i++) {
    u = q.back().first;
    v = q.back().second;
    if ((vis[u] == 1) or (vis[v] == 1)) {
      anss.push_back(an);
      q.pop_back();
      continue;
    }
    g[u].erase(v);
    g[v].erase(u);
    if (g[u].size() < k) {
      bfsq.push(u);
      vis[u] = 1;
    }
    if (g[v].size() < k) {
      bfsq.push(v);
      vis[v] = 1;
    }
    while (bfsq.size() != 0) {
      for (auto adj : g[bfsq.front()]) {
        g[adj].erase(bfsq.front());
        if (g[adj].size() < k) {
          if (vis[adj] == 0) bfsq.push(adj);
          vis[adj] = 1;
        }
      }
      ans.erase(bfsq.front());
      bfsq.pop();
      an--;
    }
    anss.push_back(an);
    q.pop_back();
  }
  anss.pop_back();
  for (int i = 0; i < m; i++) {
    printf("%lld\n", anss.back());
    anss.pop_back();
  }
  return 0;
}

```