Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 ≤ a1 ≤ a2 ≤ ... ≤ an). 

Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:

  * the bottom of the box touches the top of a stair; 
  * the bottom of the box touches the top of a box, thrown earlier. 



We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.

You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 ≤ ai ≤ 109; ai ≤ ai + 1).

The next line contains integer m (1 ≤ m ≤ 105) — the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 ≤ wi ≤ n; 1 ≤ hi ≤ 109) — the size of the i-th thrown box.

The numbers in the lines are separated by spaces.

Output

Print m integers — for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.

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

Examples

Input

5
1 2 3 6 6
4
1 1
3 1
1 1
4 3


Output

1
3
4
6


Input

3
1 2 3
2
1 1
3 1


Output

1
3


Input

1
1
5
1 2
1 10
1 10
1 10
1 10


Output

1
3
13
23
33

Note

The first sample are shown on the picture.

<image>
## necessary imports
import sys

input = sys.stdin.readline
from math import log2, log, ceil


# swap_array function
def swaparr(arr, a, b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp


## gcd function
def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


## nCr function efficient using Binomial Cofficient
def nCr(n, k):
    if (k > n - k):
        k = n - k
    res = 1
    for i in range(k):
        res = res * (n - i)
        res = res / (i + 1)
    return res


## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
    hi = len(a)
    while lo < hi:
        mid = (lo + hi) // 2
        if a[mid] < x:
            lo = mid + 1
        else:
            hi = mid
    return lo


## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while (n % 2 == 0):
        primes[2] = primes.get(2, 0) + 1
        n = n // 2
    for i in range(3, int(n ** 0.5) + 2, 2):
        while (n % i == 0):
            primes[i] = primes.get(i, 0) + 1
            n = n // i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes


## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
    res = 1
    x = x % p
    if (x == 0):
        return 0
    while (y > 0):
        if ((y & 1) == 1):
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res


## DISJOINT SET UNINON FUNCTIONS
def swap(a, b):
    temp = a
    a = b
    b = temp
    return a, b


# find function
def find(x, link):
    while (x != link[x]):
        x = link[x]
    return x


# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x, y = swap(x, y)
    if x != y:
        size[x] += size[y]
        link[y] = x


## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
    return prime


#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)

'''
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i * i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x // spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)'''
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def find_parent(parent, i,cnt):
    cnt+=1
    if parent[i-1] ==i:
        return cnt
    if parent[i-1] !=i:
        find_parent(parent, parent[i-1],cnt)

##########################################################
#for _ in range(int(input())):
# from collections import deque
from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
#n,k= map(int, input().split())
#for _ in range(int(input())):
#n,k= map(int, input().split())
from collections import Counter
import math
#for i in range(int(input())):
    #n,k=map(int, input().split())
n=int(input())
arr=list(map(int,input().split()))
m=int(input())
box=[]
for i in range(m):
    u,v=map(int,input().split())
    box.append((u,v))
ans=0
val=0
for i in range(m):
    ans=max(val+ans,arr[box[i][0]-1])
    val=box[i][1]
    print(ans)
