Smart Beaver recently got interested in a new word game. The point is as follows: count the number of distinct good substrings of some string s. To determine if a string is good or not the game uses rules. Overall there are n rules. Each rule is described by a group of three (p, l, r), where p is a string and l and r (l ≤ r) are integers. We’ll say that string t complies with rule (p, l, r), if the number of occurrences of string t in string p lies between l and r, inclusive. For example, string "ab", complies with rules ("ab", 1, 2) and ("aab", 0, 1), but does not comply with rules ("cd", 1, 2) and ("abab", 0, 1).

A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s_1s_2... s_{|}s| (|s| is a length of s) is string s_{l}s_{l} + 1... s_{r}.

Consider a number of occurrences  of string t in string p as a number of pairs of integers l, r (1 ≤ l ≤ r ≤ |p|) such that p[l... r] = t.

We’ll say that string t is good if it complies with all n rules. Smart Beaver asks you to help him to write a program that can calculate the number of distinct good substrings of string s. Two substrings s[x... y] and s[z... w] are cosidered to be distinct iff s[x... y] ≠ s[z... w].


-----Input-----

The first line contains string s. The second line contains integer n. Next n lines contain the rules, one per line. Each of these lines contains a string and two integers p_{i}, l_{i}, r_{i}, separated by single spaces (0 ≤ l_{i} ≤ r_{i} ≤ |p_{i}|). It is guaranteed that all the given strings are non-empty and only contain lowercase English letters.

The input limits for scoring 30 points are (subproblem G1):   0 ≤ n ≤ 10.  The length of string s and the maximum length of string p is  ≤ 200. 

The input limits for scoring 70 points are (subproblems G1+G2):   0 ≤ n ≤ 10.  The length of string s and the maximum length of string p is  ≤ 2000. 

The input limits for scoring 100 points are (subproblems G1+G2+G3):   0 ≤ n ≤ 10.  The length of string s and the maximum length of string p is  ≤ 50000. 


-----Output-----

Print a single integer — the number of good substrings of string s.


-----Examples-----
Input
aaab
2
aa 0 0
aab 1 1

Output
3

Input
ltntlnen
3
n 0 0
ttlneenl 1 4
lelllt 1 1

Output
2

Input
a
0

Output
1



-----Note-----

There are three good substrings in the first sample test: «aab», «ab» and «b».

In the second test only substrings «e» and «t» are good.

Source: codeforces | Difficulty: MEDIUM_HARD
Tags: String algorithms


Solutions:
from sys import stdin
import re

def substrings(s):
	for i in range(0, len(s)):
		for l in range(i, len(s)):
			yield s[i:l + 1]
test = stdin.readline().rstrip('\n')
ruleCount = int(stdin.readline())
rules = []
for i in range(0, ruleCount):
	ruleStr = stdin.readline()
	sp = ruleStr.split(' ')
	m = {}
	for s in substrings(sp[0]):
		m[s] = m.get(s, 0) + 1
	rules.append((sp[0], int(sp[1]), int(sp[2]), m))

def works(x):
	for rule in rules:
		if not rule[1] <= rule[3].get(x, 0) <= rule[2]:
			return False
	return True
print(len([x for x in set(substrings(test)) if works(x)]))


def count(p, s):
	start = 0
	c = 0
	while True:
		try:
			pos = s.index(p, start)
			c += 1
			start = pos + 1
		except ValueError:
			return c
s = input()
n = int(input())
pravs = []
for i in range(n):
	(p, l, r) = input().split()
	l = int(l)
	r = int(r)
	pravs.append((p, l, r))
var = set()
for l in range(len(s)):
	for r in range(l + 1, len(s) + 1):
		pods = s[l:r]
		for prav in pravs:
			if not prav[1] <= count(pods, prav[0]) <= prav[2]:
				break
		else:
			var.add(pods)
print(len(var))