#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = double;
using pii = pair<int, int>;
using vi = vector<int>;
const int maxn = 5e5;
const int inf = 1e9;
const int mod = 1e9 + 7;
const ll inf64 = 1e18;
const ld pi = acos(-1.0);
const ld eps = 1e-6;
int n, m, k, deg[maxn];
set<int> g[maxn];
int uu[maxn], vv[maxn];
int ans[maxn];
int del[maxn];
int cur;
void go(int v) {
  queue<int> q;
  q.push(v);
  while (!q.empty()) {
    v = q.front();
    q.pop();
    if (del[v]) continue;
    del[v] = 1;
    cur--;
    for (int to : g[v]) {
      g[to].erase(v);
      if (--deg[to] < k) q.push(to);
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].insert(v);
    g[v].insert(u);
    deg[u]++;
    deg[v]++;
    uu[i] = u;
    vv[i] = v;
  }
  cur = n;
  for (int i = 0; i < n; i++)
    if (del[i] == 0 && deg[i] < k) {
      go(i);
    }
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = cur;
    if (del[uu[i]] == 0 && del[vv[i]] == 0) {
      deg[uu[i]]--;
      deg[vv[i]]--;
      g[uu[i]].erase(vv[i]);
      g[vv[i]].erase(uu[i]);
      if (deg[uu[i]] < k) go(uu[i]);
      if (deg[vv[i]] < k) go(vv[i]);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
}

### 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 namespace std;
using ll = long long;
using ld = double;
using pii = pair<int, int>;
using vi = vector<int>;
const int maxn = 5e5;
const int inf = 1e9;
const int mod = 1e9 + 7;
const ll inf64 = 1e18;
const ld pi = acos(-1.0);
const ld eps = 1e-6;
int n, m, k, deg[maxn];
set<int> g[maxn];
int uu[maxn], vv[maxn];
int ans[maxn];
int del[maxn];
int cur;
void go(int v) {
  queue<int> q;
  q.push(v);
  while (!q.empty()) {
    v = q.front();
    q.pop();
    if (del[v]) continue;
    del[v] = 1;
    cur--;
    for (int to : g[v]) {
      g[to].erase(v);
      if (--deg[to] < k) q.push(to);
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].insert(v);
    g[v].insert(u);
    deg[u]++;
    deg[v]++;
    uu[i] = u;
    vv[i] = v;
  }
  cur = n;
  for (int i = 0; i < n; i++)
    if (del[i] == 0 && deg[i] < k) {
      go(i);
    }
  for (int i = m - 1; i >= 0; i--) {
    ans[i] = cur;
    if (del[uu[i]] == 0 && del[vv[i]] == 0) {
      deg[uu[i]]--;
      deg[vv[i]]--;
      g[uu[i]].erase(vv[i]);
      g[vv[i]].erase(uu[i]);
      if (deg[uu[i]] < k) go(uu[i]);
      if (deg[vv[i]] < k) go(vv[i]);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
}

```