#include <bits/stdc++.h>
using namespace std;
int kmp(char needle[10000], char haystack[10000]) {
  int counter = 0;
  int m = strlen(needle);
  vector<int> border(m + 1);
  border[0] = -1;
  for (__typeof(m) i = 0; i < m; i++) {
    border[i + 1] = border[i];
    while (border[i + 1] > -1 && needle[border[i + 1]] != needle[i]) {
      border[i + 1] = border[border[i + 1]];
    }
    border[i + 1]++;
  }
  int n = strlen(haystack);
  int seen = 0;
  for (__typeof(n) i = 0; i < n; i++) {
    while (seen > -1 && needle[seen] != haystack[i]) {
      seen = border[seen];
    }
    if (++seen == m) {
      counter++;
      seen = border[m];
    }
  }
  return counter;
}
char p[50000][300];
int main() {
  char s[1000], rule[10][1000], needle[1000];
  int n, l[100], r[100], ans = 0;
  scanf("%s%d", s, &n);
  for (__typeof(n) i = 0; i < n; i++) scanf("%s %d %d", rule[i], &l[i], &r[i]);
  int len = strlen(s);
  for (__typeof(len) i = 0; i < len; i++) {
    for (int j = i; j < len; j++) {
      char x = s[j + 1];
      s[j + 1] = 0;
      strcpy(needle, s + i);
      s[j + 1] = x;
      int flag = 1, curr;
      for (int k = 0; k < n && (flag); k++) {
        if (strlen(needle) > strlen(rule[k]))
          curr = 0;
        else
          curr = kmp(needle, rule[k]);
        if ((curr >= l[k] && curr <= r[k]))
          flag = 1;
        else
          flag = 0;
      }
      ans = ans + flag;
      if (flag) strcpy(p[ans - 1], needle);
    }
  }
  int c[1000] = {0};
  for (__typeof(ans) i = 0; i < ans; i++) c[kmp(p[i], s)]++;
  ans = 0;
  for (int i = 1; i < 1000; i++) ans = ans + c[i] / i;
  printf("%d\n", ans);
}

### Prompt

In cpp, your task is to solve 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 kmp(char needle[10000], char haystack[10000]) {
  int counter = 0;
  int m = strlen(needle);
  vector<int> border(m + 1);
  border[0] = -1;
  for (__typeof(m) i = 0; i < m; i++) {
    border[i + 1] = border[i];
    while (border[i + 1] > -1 && needle[border[i + 1]] != needle[i]) {
      border[i + 1] = border[border[i + 1]];
    }
    border[i + 1]++;
  }
  int n = strlen(haystack);
  int seen = 0;
  for (__typeof(n) i = 0; i < n; i++) {
    while (seen > -1 && needle[seen] != haystack[i]) {
      seen = border[seen];
    }
    if (++seen == m) {
      counter++;
      seen = border[m];
    }
  }
  return counter;
}
char p[50000][300];
int main() {
  char s[1000], rule[10][1000], needle[1000];
  int n, l[100], r[100], ans = 0;
  scanf("%s%d", s, &n);
  for (__typeof(n) i = 0; i < n; i++) scanf("%s %d %d", rule[i], &l[i], &r[i]);
  int len = strlen(s);
  for (__typeof(len) i = 0; i < len; i++) {
    for (int j = i; j < len; j++) {
      char x = s[j + 1];
      s[j + 1] = 0;
      strcpy(needle, s + i);
      s[j + 1] = x;
      int flag = 1, curr;
      for (int k = 0; k < n && (flag); k++) {
        if (strlen(needle) > strlen(rule[k]))
          curr = 0;
        else
          curr = kmp(needle, rule[k]);
        if ((curr >= l[k] && curr <= r[k]))
          flag = 1;
        else
          flag = 0;
      }
      ans = ans + flag;
      if (flag) strcpy(p[ans - 1], needle);
    }
  }
  int c[1000] = {0};
  for (__typeof(ans) i = 0; i < ans; i++) c[kmp(p[i], s)]++;
  ans = 0;
  for (int i = 1; i < 1000; i++) ans = ans + c[i] / i;
  printf("%d\n", ans);
}

```