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 = s1s2... s|s| (|s| is a length of s) is string slsl + 1... sr.

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 pi, li, ri, separated by single spaces (0 ≤ li ≤ ri ≤ |pi|). 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.
import java.io.*;
import java.util.*;
import java.math.*;
import java.awt.geom.*;

import static java.lang.Math.*;

public class Solution implements Runnable {
	
	int n, m;
	char p [][];
	char s [], a[];
	int [][] lcp;
	int [] l, r;
	
	int[] z (int n, char s []) {
		int z [] = new int [n];
		for (int i = 1, l = 0, r = 0; i < n; i++) {
			if (i <= r)
				z[i] = min (r - i + 1, z[i - l]);
			while (i + z[i] < n && s[z[i]] == s[i + z[i]])
				z[i]++;
			if (i + z[i] - 1 > r) {
				l = i;
				r = i + z[i] - 1;
			}
		}
		return z;
	}
	
	public void solve() throws Exception {
		a = sc.nextToken().toCharArray();
		
		n = sc.nextInt();
		p = new char [n][];
		l = new int [n];
		r = new int [n];
		int maxP = 0;
		for (int i = 0; i < n; i++) {
			p[i] = sc.nextToken().toCharArray();
			l[i] = sc.nextInt();
			r[i] = sc.nextInt();
			maxP = max (maxP, p[i].length);
		}
		
		m = a.length;
		int good [][] = new int [m][m + 2];
		lcp = new int [m + 1][m + 1];
		for (int i = m - 1; i >= 0; i--) {
			for (int j = m - 1; j >= 0; j--) {
				if (a[i] == a[j]) lcp[i][j] = lcp[i + 1][j + 1] + 1;
			}
		}
		
		s = new char [maxP + m];
		int count [] = new int [m + 2];
		int sum [] = new int [m + 2];
		
		for (int i = 0; i < m; i++) {
			for (int j = 1; j <= m - i; j++)
				good[i][j] = 1;
			for (int j = i; j < m; j++)
				s[j - i] = a[j];
			
			for (int k = 0; k < n; k++) {
				for (int j = 0; j < p[k].length; j++)
					s[m - i + j] = p[k][j];
				int totalLen = m - i + p[k].length;
				int z [] = z (totalLen, s);
				for (int j = 0; j <= m - i; j++)
					count[j] = 0;
				
				for (int j = m - i; j < totalLen; j++) {
					z[j] = min (z[j], m - i);
					count[z[j]]++;
				}
				
				sum[m - i + 1] = 0;
				for (int j = m - i; j >= 0; j--)
					sum[j] = sum[j + 1] + count[j];
				for (int j = 1; j <= m - i; j++)
					if (sum[j] < l[k] || sum[j] > r[k])
						good[i][j] = 0;
			}
		}
		
		Integer perm [] = new Integer [m];
		for (int i = 0; i < m; i++)
			perm[i] = i;
		
		Arrays.sort(perm, new Comparator<Integer>() {
			@Override
			public int compare(Integer arg0, Integer arg1) {
				int get = lcp[arg0][arg1];
				if (arg0 + get >= m) return -1;
				if (arg1 + get >= m) return 1;
				return Character.compare(a[arg0 + get], a[arg1 + get]);
			}
		});
		
		int result = 0;
		int bad = 0;
		for (int i = 0; i < m; i++) {
			int index = perm[i];
			for (int k = bad + 1; k <= m - index; k++)
				result += good[index][k];
			if (i + 1 < m) bad = lcp[perm[i]][perm[i + 1]];
		}
		
		out.println(result);
	}

	BufferedReader in;
	PrintWriter out;
	FastScanner sc;

	static Throwable uncaught;

	@Override
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			sc = new FastScanner(in);
			solve();
		} catch (Throwable t) {
			Solution.uncaught = t;
		} finally {
			out.close();
		}
	}

	public static void main(String[] args) throws Throwable {
		Thread t = new Thread(null, new Solution(), "", (1 << 26));
		t.start();
		t.join();
		if (uncaught != null) {
			throw uncaught;
		}
	}

}

class FastScanner {

	BufferedReader reader;
	StringTokenizer strTok;

	public FastScanner(BufferedReader reader) {
		this.reader = reader;
	}

	public String nextToken() throws IOException {
		while (strTok == null || !strTok.hasMoreTokens()) {
			strTok = new StringTokenizer(reader.readLine());
		}
		return strTok.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

}
