Sereja loves number sequences very much. That's why he decided to make himself a new one following a certain algorithm.

Sereja takes a blank piece of paper. Then he starts writing out the sequence in m stages. Each time he either adds a new number to the end of the sequence or takes l first elements of the current sequence and adds them c times to the end. More formally, if we represent the current sequence as a_1, a_2, ..., a_{n}, then after we apply the described operation, the sequence transforms into a_1, a_2, ..., a_{n}[, a_1, a_2, ..., a_{l}] (the block in the square brackets must be repeated c times). 

A day has passed and Sereja has completed the sequence. He wonders what are the values of some of its elements. Help Sereja.


-----Input-----

The first line contains integer m (1 ≤ m ≤ 10^5) — the number of stages to build a sequence. 

Next m lines contain the description of the stages in the order they follow. The first number in the line is a type of stage (1 or 2). Type 1 means adding one number to the end of the sequence, in this case the line contains integer x_{i} (1 ≤ x_{i} ≤ 10^5) — the number to add. Type 2 means copying a prefix of length l_{i} to the end c_{i} times, in this case the line further contains two integers l_{i}, c_{i} (1 ≤ l_{i} ≤ 10^5, 1 ≤ c_{i} ≤ 10^4), l_{i} is the length of the prefix, c_{i} is the number of copyings. It is guaranteed that the length of prefix l_{i} is never larger than the current length of the sequence.

The next line contains integer n (1 ≤ n ≤ 10^5) — the number of elements Sereja is interested in. The next line contains the numbers of elements of the final sequence Sereja is interested in. The numbers are given in the strictly increasing order. It is guaranteed that all numbers are strictly larger than zero and do not exceed the length of the resulting sequence. Consider the elements of the final sequence numbered starting from 1 from the beginning to the end of the sequence.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.


-----Output-----

Print the elements that Sereja is interested in, in the order in which their numbers occur in the input. 


-----Examples-----
Input
6
1 1
1 2
2 2 1
1 3
2 5 2
1 4
16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Output
1 2 1 2 3 1 2 1 2 3 1 2 1 2 3 4

```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=int(input())\na=[]\nfor i in range(n):\n    a.append(list(map(int,input().split())))\nm=int(input())\nb=list(map(lambda x:int(x)-1,input().split()))\nc=[]\nnow=0\nk=0\nans=[]\nfor i in range(n):\n    t=a[i]\n    last=now\n    if t[0]==1:\n        now+=1\n        if len(c)<100000: c.append(t[1])\n        if k<m and b[k]==now-1: \n            ans.append(t[1])\n            k+=1\n    else:\n        now+=t[1]*t[2]\n        while t[2]:\n            if len(c)<100000: c.extend(c[:t[1]])\n            else: break\n            t[2]-=1\n        while k<m and last<=b[k]<now:\n            ans.append(c[(b[k]-last)%t[1]])\n            k+=1    \nfor i in range(m):\n    print(ans[i],end=' ')\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13873