Teachers of one programming summer school decided to make a surprise for the students by giving them names in the style of the "Hobbit" movie. Each student must get a pseudonym maximally similar to his own name. The pseudonym must be a name of some character of the popular saga and now the teachers are busy matching pseudonyms to student names.

There are n students in a summer school. Teachers chose exactly n pseudonyms for them. Each student must get exactly one pseudonym corresponding to him. Let us determine the relevance of a pseudonym b to a student with name a as the length of the largest common prefix a and b. We will represent such value as $\operatorname{lcp}(a, b)$. Then we can determine the quality of matching of the pseudonyms to students as a sum of relevances of all pseudonyms to the corresponding students.

Find the matching between students and pseudonyms with the maximum quality.


-----Input-----

The first line contains number n (1 ≤ n ≤ 100 000) — the number of students in the summer school.

Next n lines contain the name of the students. Each name is a non-empty word consisting of lowercase English letters. Some names can be repeating.

The last n lines contain the given pseudonyms. Each pseudonym is a non-empty word consisting of small English letters. Some pseudonyms can be repeating.

The total length of all the names and pseudonyms doesn't exceed 800 000 characters.


-----Output-----

In the first line print the maximum possible quality of matching pseudonyms to students.

In the next n lines describe the optimal matching. Each line must have the form a b (1 ≤ a, b ≤ n), that means that the student who was number a in the input, must match to the pseudonym number b in the input.

The matching should be a one-to-one correspondence, that is, each student and each pseudonym should occur exactly once in your output. If there are several optimal answers, output any.


-----Examples-----
Input
5
gennady
galya
boris
bill
toshik
bilbo
torin
gendalf
smaug
galadriel

Output
11
4 1
2 5
1 3
5 2
3 4



-----Note-----

The first test from the statement the match looks as follows:   bill  →  bilbo (lcp = 3)  galya  →  galadriel (lcp = 3)  gennady  →  gendalf (lcp = 3)  toshik  →  torin (lcp = 2)  boris  →  smaug (lcp = 0)

Difficulty: HARD
Tags: Tree algorithms

Graph traversal

String algorithms

Starter code:

Solutions:
import sys

class Node:

	def __init__(self, d):
		global nodes
		self.ch = {}
		self.a = [[], []]
		self.d = d
		nodes += [self]
nodes = []
pairs = []
res = 0
N = int(sys.stdin.readline())
_input = sys.stdin.readlines()
_input = [s[:-1] for s in _input]
A = [_input[:N], _input[N:]]
T = Node(0)
for (i, l) in enumerate(A):
	for (j, s) in enumerate(l):
		t = T
		for c in s:
			v = ord(c) - ord('a')
			if not v in t.ch:
				t.ch[v] = Node(t.d + 1)
			t = t.ch[v]
		t.a[i] += [j + 1]
for n in reversed(nodes):
	for i in n.ch:
		n.a[0] += n.ch[i].a[0]
		n.a[1] += n.ch[i].a[1]
	k = min(len(n.a[0]), len(n.a[1]))
	for i in range(k):
		pairs += [str(n.a[0][-1]) + ' ' + str(n.a[1][-1])]
		n.a[0].pop()
		n.a[1].pop()
		res += n.d
print(res)
print('\n'.join(pairs))


import sys
SIGMA = 26
nodes = []
pairs = []
res = 0

class Node:

	def __init__(self):
		self.ch = {}
		self.a = []
		self.b = []
		self.d = 0

	def add(self, s, i):
		t = self
		for c in s:
			v = ord(c) - ord('a')
			if not v in t.ch:
				t.ch[v] = Node()
				t.ch[v].d = t.d + 1
				nodes.append(t.ch[v])
			t = t.ch[v]
		t.a.append(i)

	def inc(self, s, i):
		t = self
		for c in s:
			v = ord(c) - ord('a')
			if not v in t.ch:
				break
			t = t.ch[v]
		t.b.append(i)

	def solve(self):
		global pairs
		global res
		for i in range(SIGMA):
			if i in self.ch:
				self.a.extend(self.ch[i].a)
				self.b.extend(self.ch[i].b)
		k = min(len(self.a), len(self.b))
		for i in range(k):
			pairs.append(str(self.a[-1]) + ' ' + str(self.b[-1]))
			self.a.pop()
			self.b.pop()
			res += self.d
		return res
sys.setrecursionlimit(2000000)
_input = sys.stdin.readlines()
_input = [s[:-1] for s in _input]
N = int(_input[0])
A = _input[1:N + 1]
B = _input[N + 1:]
T = Node()
nodes.append(T)
for (i, s) in enumerate(A):
	T.add(s, i + 1)
for (i, s) in enumerate(B):
	T.inc(s, i + 1)
for n in reversed(nodes):
	n.solve()
print(res)
print('\n'.join(pairs))