#include <bits/stdc++.h>
using namespace std;
int main() {
  set<string> m;
  int ans[1000000];
  string s, s1, s2;
  cin >> s;
  for (int i = 1; i <= s.size(); i++) {
    for (int j = 0; j < s.size() - i + 1; j++) {
      s1 = s.substr(j, i);
      m.insert(s1);
    }
  }
  int n, k, k1, p, z, br, ans1 = 0, l;
  cin >> n;
  if (n == 0) {
    cout << m.size() << "\n";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    z = 0;
    cin >> s1;
    cin >> k >> k1;
    for (set<string>::iterator it = m.begin(); it != m.end(); it++) {
      br = 1;
      s2 = *it;
      p = 0;
      l = s1.find(s2, 0);
      p = l + 1;
      while (l != string::npos) {
        l = s1.find(s2, p);
        p = l + 1;
        br++;
        if (br > k1 + 1) break;
      }
      br--;
      if (br >= k && br <= k1) {
        ans[z]++;
        if (i + 1 == n && ans[z] == n) {
          ans1++;
        }
      }
      z++;
    }
  }
  cout << ans1 << "\n";
  return 0;
}

### 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 main() {
  set<string> m;
  int ans[1000000];
  string s, s1, s2;
  cin >> s;
  for (int i = 1; i <= s.size(); i++) {
    for (int j = 0; j < s.size() - i + 1; j++) {
      s1 = s.substr(j, i);
      m.insert(s1);
    }
  }
  int n, k, k1, p, z, br, ans1 = 0, l;
  cin >> n;
  if (n == 0) {
    cout << m.size() << "\n";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    z = 0;
    cin >> s1;
    cin >> k >> k1;
    for (set<string>::iterator it = m.begin(); it != m.end(); it++) {
      br = 1;
      s2 = *it;
      p = 0;
      l = s1.find(s2, 0);
      p = l + 1;
      while (l != string::npos) {
        l = s1.find(s2, p);
        p = l + 1;
        br++;
        if (br > k1 + 1) break;
      }
      br--;
      if (br >= k && br <= k1) {
        ans[z]++;
        if (i + 1 == n && ans[z] == n) {
          ans1++;
        }
      }
      z++;
    }
  }
  cout << ans1 << "\n";
  return 0;
}

```