#include <bits/stdc++.h>
using namespace std;
const int max6 = 1e6 + 6;
const int oo = 2e9 + 9;
const long long inf = 2e18 + 18;
vector<pair<int, int> > g[max6];
pair<int, int> a[max6];
int q[max6];
int block[max6];
int deg[max6];
int del[max6];
int res[max6];
int n, m, need;
int main() {
  cin >> n >> m >> need;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back({v, i});
    g[v].push_back({u, i});
    a[i] = {u, v};
    deg[u]++, deg[v]++;
  }
  int l = 1, r = 0;
  for (int i = 1; i <= n; ++i)
    if (deg[i] < need) q[++r] = i, del[i] = 1;
  int maxMem = n;
  for (int i = m; i >= 1; --i) {
    while (l <= r) {
      int u = q[l++];
      maxMem--;
      for (auto it : g[u]) {
        int id = it.second;
        int v = it.first;
        if (block[id]) continue;
        if (del[v]) continue;
        block[id] = 1;
        deg[v]--;
        if (deg[v] < need) {
          del[v] = 1;
          q[++r] = v;
        }
      }
    }
    res[i] = maxMem;
    int u = a[i].first;
    int v = a[i].second;
    if (del[u] || del[v]) continue;
    deg[u]--;
    deg[v]--;
    block[i] = 1;
    if (deg[u] < need) {
      q[++r] = u;
      del[u] = 1;
    }
    if (deg[v] < need) {
      q[++r] = v;
      del[v] = 1;
    }
  }
  for (int i = 1; i <= m; ++i) cout << res[i] << "\n";
}

### Prompt

Create a solution in cpp for 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 max6 = 1e6 + 6;
const int oo = 2e9 + 9;
const long long inf = 2e18 + 18;
vector<pair<int, int> > g[max6];
pair<int, int> a[max6];
int q[max6];
int block[max6];
int deg[max6];
int del[max6];
int res[max6];
int n, m, need;
int main() {
  cin >> n >> m >> need;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back({v, i});
    g[v].push_back({u, i});
    a[i] = {u, v};
    deg[u]++, deg[v]++;
  }
  int l = 1, r = 0;
  for (int i = 1; i <= n; ++i)
    if (deg[i] < need) q[++r] = i, del[i] = 1;
  int maxMem = n;
  for (int i = m; i >= 1; --i) {
    while (l <= r) {
      int u = q[l++];
      maxMem--;
      for (auto it : g[u]) {
        int id = it.second;
        int v = it.first;
        if (block[id]) continue;
        if (del[v]) continue;
        block[id] = 1;
        deg[v]--;
        if (deg[v] < need) {
          del[v] = 1;
          q[++r] = v;
        }
      }
    }
    res[i] = maxMem;
    int u = a[i].first;
    int v = a[i].second;
    if (del[u] || del[v]) continue;
    deg[u]--;
    deg[v]--;
    block[i] = 1;
    if (deg[u] < need) {
      q[++r] = u;
      del[u] = 1;
    }
    if (deg[v] < need) {
      q[++r] = v;
      del[v] = 1;
    }
  }
  for (int i = 1; i <= m; ++i) cout << res[i] << "\n";
}

```