Problem: The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.

The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.

Input

The first line contains a single integer n (2 ≤ n ≤ 500) — the number of lower-level employees.

This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 ≤ a_{i,j} ≤ 5000) — salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.

Output

In the first line, print a single integer k — the number of employees in the company.

In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.

In the third line, print a single integer r — the number of the employee who is the head of the company.

In the following k-1 lines, print two integers v and u (1 ≤ v, u ≤ k) — the number of the employee and his direct supervisor.

Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.

Example

Input


3
2 5 7
5 1 7
7 7 4


Output


5
2 1 4 7 5 
4
1 5
2 5
5 4
3 4

Note

One of the possible structures in the first example: <image>

Solution (in PYTHON3):

Solution:
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""

import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        import os
        self.os = os
        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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
            self.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")

#start_time = time.time()

def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'

MOD = 10**9 + 7 

"""
We can map out the ancestral path for i == 0
Then for each subsequent i, we find the lowest point it joins an existing path, and we then create a path up to that point
"""


def solve():
    N = getInt()
    M = getMat(N)
    parent = [-1]*N
    salary = [-1]*N
    lookup = dd(int) # lookup[(i,s)] is the index of the node corresponding to the ancestor of i with salary s
    for i in range(N):
        salary[i] = M[i][i]
        to_join = -1
        lowest = 10**4
        for j in range(i):
            if M[i][j] < lowest:
                lowest = M[i][j]
                to_join = j
        #so we want to join the path of node to_join at its value _lowest_
        if i == 0:
            prev = 0
            B = sorted([(a,j) for j,a in enumerate(M[0])])
            for j in range(1,N):
                if B[j][0] == B[j-1][0]:
                    lookup[(i,s)] = prev
                else:
                    idx = len(parent)
                    s = B[j][0]
                    lookup[(i,s)] = idx
                    parent.append(-1)
                    parent[prev] = idx
                    salary.append(s)
                    prev = idx
        else:
            assert(to_join >= 0)
            prev = i
            flag = False
            B = sorted([(a,j) for j,a in enumerate(M[i])])
            for j in range(1,N):
                if B[j][0] == lowest:
                    idx = lookup[(to_join,lowest)]
                    lookup[(i,lowest)] = idx
                    parent[prev] = idx
                    flag = True
                elif B[j][0] != B[j-1][0] and not flag:
                    idx = len(parent)
                    s = B[j][0]
                    lookup[(i,s)] = idx
                    parent.append(-1)
                    parent[prev] = idx
                    salary.append(s)
                    prev = idx
                elif not flag:
                    lookup[(i,s)] = prev
                else:
                    lookup[(i,s)] == lookup[(to_join,s)]
                    
                    
    print(len(parent))
    print(*salary)
    for i,p in enumerate(parent):
        if p == -1: print(i+1)
    for i,p in enumerate(parent):
        if p == -1: continue
        print(i+1,p+1)
    return
    
#for _ in range(getInt()):
#print(solve())
solve()

#print(time.time()-start_time)