Problem: It's Petya's birthday party and his friends have presented him a brand new "Electrician-n" construction set, which they are sure he will enjoy as he always does with weird puzzles they give him.

Construction set "Electrician-n" consists of 2n - 1 wires and 2n light bulbs. Each bulb has its own unique index that is an integer from 1 to 2n, while all wires look the same and are indistinguishable. In order to complete this construction set one has to use each of the wires to connect two distinct bulbs. We define a chain in a completed construction set as a sequence of distinct bulbs of length at least two, such that every two consecutive bulbs in this sequence are directly connected by a wire. Completed construction set configuration is said to be correct if a resulting network of bulbs and wires has a tree structure, i.e. any two distinct bulbs are the endpoints of some chain.

Petya was assembling different configurations for several days, and he noticed that sometimes some of the bulbs turn on. After a series of experiments he came up with a conclusion that bulbs indexed 2i and 2i - 1 turn on if the chain connecting them consists of exactly d_i wires. Moreover, the following important condition holds: the value of d_i is never greater than n.

Petya did his best but was not able to find a configuration that makes all bulbs to turn on, so he seeks your assistance. Please, find out a configuration that makes all bulbs shine. It is guaranteed that such configuration always exists.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the parameter of a construction set that defines the number of bulbs and the number of wires.

Next line contains n integers d_1, d_2, …, d_n (1 ≤ d_i ≤ n), where d_i stands for the number of wires the chain between bulbs 2i and 2i - 1 should consist of.

Output

Print 2n - 1 lines. The i-th of them should contain two distinct integers a_i and b_i (1 ≤ a_i, b_i ≤ 2n, a_i ≠ b_i) — indices of bulbs connected by a wire.

If there are several possible valid answer you can print any of them.

Examples

Input


3
2 2 2


Output


1 6
2 6
3 5
3 6
4 5


Input


4
2 2 2 1


Output


1 6
1 7
2 6
3 5
3 6
4 5
7 8


Input


6
2 2 2 2 2 2


Output


1 3
2 3
3 5
4 5
5 7
6 7
7 12
8 12
9 11
9 12
10 11


Input


2
1 1


Output


1 2
1 4
3 4

Note

<image> Answer for the first sample test.  <image> Answer for the second sample test. 

Solution (in PYTHON):

Solution:
from __future__ import division, print_function
def main():
    n=int(input())
    d=list(map(int,input().split()))
    temp=[]
    for i in range(n):
        temp.append((d[i],i+1))
    temp.sort()
    temp=temp[::-1]
    x=[0]*n
    pos={}
    for i in range(n):
        x[i]=2*temp[i][1]-1
        pos[2*temp[i][1]-1]=i
    edges={}
    for item in temp:
        if (pos[2*item[1]-1]+item[0])==len(x):
            x.append(item[1]*2)
        else :
            y=pos[2*item[1]-1]+item[0]-1
            if x[y] not in edges:
                edges[x[y]]=[2*item[1]]
            else :
                edges[x[y]].append(2*item[1])
    for i in range(len(x)-1):
        print(x[i],x[i+1])
    for k in edges:
        for item in edges[k]:
            print(k,item)
        
        
    
        
    


######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

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

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')

# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'

# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

if __name__== "__main__":
  main()