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.
#include <bits/stdc++.h>
using namespace std;
int w[41000], h[41000], sa[41000], r[41000], wv[41000], t[41000], ans, i, j, k,
    X[12], Y[12], num[12], n, tr[41000], L, m, l, be[41000], x, y;
char a[41000], ch[41000];
void da() {
  int i, m, j, p;
  memset(r, 0, sizeof(r));
  memset(w, 0, sizeof(w));
  for (i = 1; i <= n; ++i) w[r[i] = a[i]]++;
  for (i = 1; i <= 2000; ++i) w[i] += w[i - 1];
  for (m = 2000, i = n; i >= 1; --i) sa[w[r[i]]--] = i;
  for (j = 1; j < n; j *= 2) {
    for (p = 0, i = n - j + 1; i <= n; ++i) tr[++p] = i;
    for (i = 1; i <= n; ++i)
      if (sa[i] > j) tr[++p] = sa[i] - j;
    for (i = 1; i <= n; ++i) wv[i] = r[tr[i]];
    for (i = 1; i <= m; ++i) w[i] = 0;
    for (i = 1; i <= n; ++i) w[wv[i]]++;
    for (i = 1; i <= m; ++i) w[i] += w[i - 1];
    for (i = n; i >= 1; --i) sa[w[wv[i]]--] = tr[i];
    for (i = 1; i <= n; ++i) t[i] = r[i];
    for (p = 0, i = 1; i <= n; ++i)
      if (t[sa[i]] == t[sa[i - 1]] && t[sa[i] + j] == t[sa[i - 1] + j])
        r[sa[i]] = p;
      else
        r[sa[i]] = ++p;
    m = p;
    if (p == n) break;
  }
}
void geth() {
  int i, j, p = 0;
  for (i = 1; i <= n; h[r[i++]] = p)
    for (p ? p-- : 0, j = sa[r[i] - 1]; a[j + p] == a[i + p]; p++)
      ;
}
int main() {
  scanf("%s", a + 1);
  n = L = strlen(a + 1);
  scanf("%d", &m);
  for (i = 1; i <= m; ++i) {
    scanf("%s", ch + 1);
    l = strlen(ch + 1);
    a[++n] = i;
    be[n] = 11;
    for (j = 1; j <= l; ++j) a[++n] = ch[j], be[n] = i;
    scanf("%d%d", &X[i], &Y[i]);
  }
  da();
  geth();
  for (i = 1; i <= L; ++i) {
    x = y = r[i];
    memset(num, 0, sizeof(num));
    for (j = L; j >= i; --j) {
      while (x - 1 >= 1 && h[x] >= j - i + 1) {
        x--;
        num[be[sa[x]]]++;
        if (be[sa[x]] == 0 && sa[x] < i) goto a1;
      }
      while (y + 1 <= n && h[y + 1] >= j - i + 1) {
        y++;
        num[be[sa[y]]]++;
        if (be[sa[y]] == 0 && sa[y] < i) goto a1;
      }
      for (k = 1; k <= m; ++k) {
        if (num[k] > Y[k]) goto a1;
        if (num[k] < X[k]) break;
      }
      if (k == m + 1) ans++;
    }
  a1:;
  }
  printf("%d\n", ans);
}
