Konrad is a Human Relations consultant working for VoltModder, a large electrical equipment producer. Today, he has been tasked with evaluating the level of happiness in the company.

There are $n$ people working for VoltModder, numbered from $1$ to $n$. Each employee earns a different amount of money in the company — initially, the $i$-th person earns $i$ rubles per day.

On each of $q$ following days, the salaries will be revised. At the end of the $i$-th day, employee $v_i$ will start earning $n+i$ rubles per day and will become the best-paid person in the company. The employee will keep his new salary until it gets revised again.

Some pairs of people don't like each other. This creates a great psychological danger in the company. Formally, if two people $a$ and $b$ dislike each other and $a$ earns more money than $b$, employee $a$ will brag about this to $b$. A dangerous triple is a triple of three employees $a$, $b$ and $c$, such that $a$ brags to $b$, who in turn brags to $c$. If $a$ dislikes $b$, then $b$ dislikes $a$.

At the beginning of each day, Konrad needs to evaluate the number of dangerous triples in the company. Can you help him do it?


-----Input-----

The first line contains two integers $n$ and $m$ ($1 \le n \le 100\,000$, $0 \le m \le 100\,000$) — the number of employees in the company and the number of pairs of people who don't like each other. Each of the following $m$ lines contains two integers $a_i$, $b_i$ ($1 \le a_i, b_i \le n$, $a_i \neq b_i$) denoting that employees $a_i$ and $b_i$ hate each other (that is, $a_i$ dislikes $b_i$ and $b_i$ dislikes $a_i$). Each such relationship will be mentioned exactly once.

The next line contains an integer $q$ ($0 \le q \le 100\,000$) — the number of salary revisions. The $i$-th of the following $q$ lines contains a single integer $v_i$ ($1 \le v_i \le n$) denoting that at the end of the $i$-th day, employee $v_i$ will earn the most.


-----Output-----

Output $q + 1$ integers. The $i$-th of them should contain the number of dangerous triples in the company at the beginning of the $i$-th day.


-----Examples-----
Input
4 5
1 2
2 4
1 3
3 4
2 3
2
2
3

Output
4
3
2

Input
3 3
1 2
2 3
1 3
5
1
2
2
1
3

Output
1
1
1
1
1
1



-----Note-----

Consider the first sample test. The $i$-th row in the following image shows the structure of the company at the beginning of the $i$-th day. A directed edge from $a$ to $b$ denotes that employee $a$ brags to employee $b$. The dangerous triples are marked by highlighted edges. [Image]

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'n, m = map(int, input().split())\nout, inp =[0]*(n+1), [0]*(n+1)\ninset = [[] for i in range(n+1)]\n\nfor i in range(m):\n    x, y = map(int, input().split())\n    \n    if x > y:\n        z = x\n        x = y\n        y = z\n    \n    out[y] += 1\n    inp[x] += 1\n    inset[x].append(y)\n\nq = int(input())\n\nres = 0\nfor ind in range(1, n+1):\n    res += inp[ind] * out[ind]\n\nprint(res)\nfor i in range(q):\n    best = int(input())\n    \n    res -= out[best] * inp[best]\n    for pred_ind in inset[best]:\n        res += -out[pred_ind] * inp[pred_ind] + ((out[pred_ind] - 1) * (inp[pred_ind] + 1))\n        out[pred_ind] -= 1\n        inp[pred_ind] += 1\n        inset[pred_ind].append(best)\n    \n    \n    out[best] += inp[best]\n    inp[best] = 0\n    inset[best] = []\n    \n    print(res)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13713