#include <bits/stdc++.h>
using namespace std;
string s, p[11], t;
pair<int, int> lim[11];
int n;
vector<string> ans;
int occ(string a, string b) {
  if (a.length() > b.length()) return 0;
  int res = 0;
  for (int i = 0; i < (b.length() - a.length() + 1); ++i)
    if (b.compare(i, a.length(), a) == 0) ++res;
  return res;
}
int main() {
  cin >> s;
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) cin >> p[i] >> lim[i].first >> lim[i].second;
  for (int len = (1); len <= (s.length()); ++len) {
    for (int start = (0); start <= (s.length() - len); ++start) {
      t = s.substr(start, len);
      bool good = true;
      for (int i = 0; i < (n); ++i) {
        int occurence = occ(t, p[i]);
        if (occurence < lim[i].first || lim[i].second < occurence) {
          good = false;
          break;
        }
      }
      if (good) ans.push_back(t);
    }
  }
  sort((ans).begin(), (ans).end());
  int nans = 0;
  for (int i = 0; i < (((int)(ans).size())); ++i)
    if (i == 0)
      nans++;
    else if (ans[i].compare(ans[i - 1]))
      nans++;
  printf("%d\n", nans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
string s, p[11], t;
pair<int, int> lim[11];
int n;
vector<string> ans;
int occ(string a, string b) {
  if (a.length() > b.length()) return 0;
  int res = 0;
  for (int i = 0; i < (b.length() - a.length() + 1); ++i)
    if (b.compare(i, a.length(), a) == 0) ++res;
  return res;
}
int main() {
  cin >> s;
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) cin >> p[i] >> lim[i].first >> lim[i].second;
  for (int len = (1); len <= (s.length()); ++len) {
    for (int start = (0); start <= (s.length() - len); ++start) {
      t = s.substr(start, len);
      bool good = true;
      for (int i = 0; i < (n); ++i) {
        int occurence = occ(t, p[i]);
        if (occurence < lim[i].first || lim[i].second < occurence) {
          good = false;
          break;
        }
      }
      if (good) ans.push_back(t);
    }
  }
  sort((ans).begin(), (ans).end());
  int nans = 0;
  for (int i = 0; i < (((int)(ans).size())); ++i)
    if (i == 0)
      nans++;
    else if (ans[i].compare(ans[i - 1]))
      nans++;
  printf("%d\n", nans);
  return 0;
}

```