#include <bits/stdc++.h>
using namespace std;
int nxt[210];
int getnum(char *s1, char *s2, int l) {
  int len1 = strlen(s1), ans = 0, len2 = strlen(s2), i, j;
  for (i = 1, j = -1, nxt[0] = -1; i < len2; i++) {
    while (j + 1 && s2[i] - s2[j + 1]) j = nxt[j];
    if (s2[i] == s2[j + 1]) j++;
    nxt[i] = j;
  }
  for (i = 0, j = -1; i < len1; i++) {
    while (j + 1 && s1[i] - s2[j + 1]) j = nxt[j];
    if (s1[i] == s2[j + 1]) j++;
    if (j == len2 - 1) ans++;
  }
  return ans;
}
char s[20][1000];
int l[1100], r[1100];
int gao(int n, char *ss, int len) {
  int i, j;
  for (i = 1; i <= n; i++) {
    j = getnum(s[i], ss, len);
    if (j < l[i] || j > r[i]) return 0;
  }
  return 1;
}
int ll[1100], rr[1100];
int ch[21000][26], cnt;
char ss[210];
char tmp[210];
int main() {
  scanf("%s", ss);
  int i, j, n = 0, k;
  int m;
  cin >> m;
  int ans = 0;
  for (i = 1; i <= m; i++) scanf("%s%d%d", s[i], &l[i], &r[i]);
  for (i = 0; ss[i]; i++) {
    n = 0;
    for (j = i; ss[j]; j++) {
      if (!ch[n][ss[j] - 'a']) {
        ch[n][ss[j] - 'a'] = ++cnt;
        for (k = i; k <= j; k++) tmp[k - i] = ss[k];
        tmp[k - i] = 0;
        ans += gao(m, tmp, j - i + 1);
      }
      n = ch[n][ss[j] - 'a'];
    }
  }
  cout << ans << endl;
}

### Prompt

Generate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int nxt[210];
int getnum(char *s1, char *s2, int l) {
  int len1 = strlen(s1), ans = 0, len2 = strlen(s2), i, j;
  for (i = 1, j = -1, nxt[0] = -1; i < len2; i++) {
    while (j + 1 && s2[i] - s2[j + 1]) j = nxt[j];
    if (s2[i] == s2[j + 1]) j++;
    nxt[i] = j;
  }
  for (i = 0, j = -1; i < len1; i++) {
    while (j + 1 && s1[i] - s2[j + 1]) j = nxt[j];
    if (s1[i] == s2[j + 1]) j++;
    if (j == len2 - 1) ans++;
  }
  return ans;
}
char s[20][1000];
int l[1100], r[1100];
int gao(int n, char *ss, int len) {
  int i, j;
  for (i = 1; i <= n; i++) {
    j = getnum(s[i], ss, len);
    if (j < l[i] || j > r[i]) return 0;
  }
  return 1;
}
int ll[1100], rr[1100];
int ch[21000][26], cnt;
char ss[210];
char tmp[210];
int main() {
  scanf("%s", ss);
  int i, j, n = 0, k;
  int m;
  cin >> m;
  int ans = 0;
  for (i = 1; i <= m; i++) scanf("%s%d%d", s[i], &l[i], &r[i]);
  for (i = 0; ss[i]; i++) {
    n = 0;
    for (j = i; ss[j]; j++) {
      if (!ch[n][ss[j] - 'a']) {
        ch[n][ss[j] - 'a'] = ++cnt;
        for (k = i; k <= j; k++) tmp[k - i] = ss[k];
        tmp[k - i] = 0;
        ans += gao(m, tmp, j - i + 1);
      }
      n = ch[n][ss[j] - 'a'];
    }
  }
  cout << ans << endl;
}

```