---
dataset_info:
  features:
  - name: name
    dtype: string
  - name: description
    dtype: string
  - name: solutions
    dtype: string
  splits:
  - name: train
    num_bytes: 22782308
    num_examples: 8139
  download_size: 10785367
  dataset_size: 22782308
configs:
- config_name: default
  data_files:
  - split: train
    path: data/train-*
task_categories:
- text-generation
language:
- en
tags:
- code
---

This dataset is based on the "code_contests" dataset by DeepMind, licensed under CC BY 4.0. 
Original dataset available at: https://huggingface.co/datasets/deepmind/code_contests



このデータセットは、[deepmind/code_contests](https://huggingface.co/datasets/deepmind/code_contests)からAtCoderのデータのみを抽出し、Pythonで正解として提出された最初のコードを取得して格納したものです。これにより、教師あり学習に適した形に整えられています。

This dataset is created by extracting only the data from AtCoder within [deepmind/code_contests](https://huggingface.co/datasets/deepmind/code_contests) and retrieving the first Python code submission marked as correct. As a result, it has been formatted to be suitable for supervised learning.

```
Dataset({
    features: ['name', 'description', 'solutions'],
    num_rows: 8139
})
```

# First element example
description
```
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.
```
solutions
```
```python
from collections import deque

def solve(adj, m, k, uv):
    n = len(adj)
    nn = [len(a) for a in adj]
    q = deque()
    for i in range(n):
        if nn[i] < k:
            q.append(i)
    while q:
        v = q.popleft()
        for u in adj[v]:
            nn[u] -= 1
            if nn[u] == k-1:
                q.append(u)
    res = [0]*m
    nk = len([1 for i in nn if i >= k])
    res[-1] = nk
    for i in range(m-1, 0, -1):
        u1, v1 = uv[i]

        if nn[u1] < k or nn[v1] < k:
            res[i - 1] = nk
            continue
        if nn[u1] == k:
            q.append(u1)
            nn[u1] -= 1
        if not q and nn[v1] == k:
            q.append(v1)
            nn[v1] -= 1

        if not q:
            nn[u1] -= 1
            nn[v1] -= 1
            adj[u1].remove(v1)
            adj[v1].remove(u1)

        while q:
            v = q.popleft()
            nk -= 1
            for u in adj[v]:
                nn[u] -= 1
                if nn[u] == k - 1:
                    q.append(u)
        res[i - 1] = nk
    return res

n, m, k = map(int, input().split())
a = [set() for i in range(n)]
uv = []
for i in range(m):
    u, v = map(int, input().split())
    a[u - 1].add(v - 1)
    a[v - 1].add(u - 1)
    uv.append((u-1, v-1))

res = solve(a, m, k, uv)
print(str(res)[1:-1].replace(' ', '').replace(',', '\n'))
```       　          
```