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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Locale;
import java.util.StringTokenizer;

public class G {
	
	private static final int MAX = 5000;
	
	private char[] S = new char[MAX];
	private int[] pref = new int[MAX];
	private int count(char[] s, char[] t, int l, int r) {
		int k = 0;
		for (int i = l; i <= r; i++) {
			S[k++] = t[i];
		}
		S[k++] = '#';
		for (char c: s) {
			S[k++] = c;
		}
		
		int len = r - l + 1;
		int res = 0;
		for (int i = 1; i < k; i++) {
			int j = pref[i - 1];
			while (j > 0 && S[i] != S[j]) {
				j = pref[j - 1];
			}
			if (S[i] == S[j]) {
				j++;
			}
			pref[i] = j;
			if (j == len) {
				res++;
			}
		}
		
		return res;
	}

	private void solve() throws IOException {
		char[] s = nextToken().toCharArray();
		int n = nextInt();
		char[][] p = new char[n][];
		int[] l = new int[n];
		int[] r = new int[n];
		
		for (int i = 0; i < n; i++) {
			p[i] = nextToken().toCharArray();
			l[i] = nextInt();
			r[i] = nextInt();
		}
		
		HashSet<Long> set = new HashSet<Long>();

		if (n == 0) {
			for (int i = 0; i < s.length; i++) {
				long h = 0;
				for (int j = i; j < s.length; j++) {
					h = h * 113 + s[j] - 'a' + 1;
					set.add(h);
				}
			}
			println(set.size());
			return;
		}
		
		
		int L = 0;
		int R = 0;
		for (int i = 0; i < s.length; i++) {
			if (L == s.length) {
				break;
			}
			if (R == s.length) {
				R--;
			}
			L = Math.max(L, i);
			R = Math.max(R, L);
			
			boolean ok = false;
			while (true) {
//				System.err.println("count L");
				boolean less = false;
				boolean more = false;
				for (int j = 0; j < n; j++) {
					int k = count(p[j], s, i, L);
					if (k < l[j]) {
						less = true;
						break;
					} else if (k > r[j]) {
						more = true;
					}
				}
				
				if (less) {
					break;
				}
				if (more) {
					L++;
					if (L == s.length) {
						break;
					}
				}
				if (!less && !more) {
					ok = true;
					break;
				}
			}
			
			if (!ok) {
				continue;
			}
			
			ok = false;
			while (true) {
//				System.err.println("count L");
				boolean less = false;
				boolean more = false;
				for (int j = 0; j < n; j++) {
					int k = count(p[j], s, i, R);
					if (k < l[j]) {
						less = true;
						break;
					} else if (k > r[j]) {
						more = true;
					}
				}
				
				if (less) {
					break;
				}
				if (more) {
					R++;
				}
				if (!less && !more) {
					ok = true;
					R++;
				}
				if (R == s.length) {
					break;
				}
			}
			
			if (!ok) {
				continue;
			}
			
			R--;
//			System.err.println(i + " " + L + " " + R);
			long h = 0;
			for (int j = i; j <= R; j++) {
				h = h * 113 + s[j] - 'a' + 1;
				if (j >= L) {
					set.add(h);
				}
			}
		}
		
		println(set.size());
	}

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

	private int nextInt() throws NumberFormatException, IOException {
		return Integer.parseInt(nextToken());
	}

	private double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(nextToken());
	}

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

	private void print(Object o) {
		writer.print(o);
	}

	private void println(Object o) {
		writer.println(o);
	}

	private void printf(String format, Object... o) {
		writer.printf(format, o);
	}

	public static void main(String[] args) {
		long time = System.currentTimeMillis();
		Locale.setDefault(Locale.US);
		new G().run();
		System.err.printf("%.3f\n", 1e-3 * (System.currentTimeMillis() - time));
	}

	BufferedReader reader;
	StringTokenizer tokenizer;
	PrintWriter writer;

	private void run() {
		try {
			reader = new BufferedReader(new InputStreamReader(System.in));
			writer = new PrintWriter(System.out);
			solve();
			reader.close();
			writer.close();
		} catch (IOException e) {
			e.printStackTrace();
			System.exit(13);
		}
	}
}