#include <bits/stdc++.h>
using namespace std;
int lens[410];
string s;
set<string> ans;
void Zfunction() {
  int n = s.size();
  int L = 0, R = 0;
  for (int i = 1; i < n; i++) {
    if (i > R) {
      L = R = i;
      while (R < n && s[R - L] == s[R]) R++;
      lens[i] = R - L;
      R--;
    } else {
      if (lens[i - L] < R - i + 1)
        lens[i] = lens[i - L];
      else {
        L = i;
        while (R < n && s[R - L] == s[R]) R++;
        lens[i] = R - L;
        R--;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string t;
  cin >> t;
  int n;
  cin >> n;
  string l[n];
  pair<int, int> p[n];
  for (int i = 0; i < n; i++) {
    cin >> l[i];
    int x, y;
    cin >> x >> y;
    p[i] = {x, y};
  }
  int res = 0;
  for (int i = 0; i < (int)t.size(); i++) {
    for (int j = i; j < (int)t.size(); j++) {
      string par = "";
      for (int k = i; k <= j; k++) par += t[k];
      bool ok = true;
      for (int k = 0; k < n; k++) {
        s = par + "#" + l[k];
        Zfunction();
        int cont = 0;
        for (int u = 0; u < (int)s.size(); u++) {
          if (lens[u] >= (int)par.size()) cont++;
        }
        if (cont < p[k].first or cont > p[k].second) ok = false;
      }
      if (ok) ans.insert(par);
    }
  }
  cout << ans.size() << 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 lens[410];
string s;
set<string> ans;
void Zfunction() {
  int n = s.size();
  int L = 0, R = 0;
  for (int i = 1; i < n; i++) {
    if (i > R) {
      L = R = i;
      while (R < n && s[R - L] == s[R]) R++;
      lens[i] = R - L;
      R--;
    } else {
      if (lens[i - L] < R - i + 1)
        lens[i] = lens[i - L];
      else {
        L = i;
        while (R < n && s[R - L] == s[R]) R++;
        lens[i] = R - L;
        R--;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string t;
  cin >> t;
  int n;
  cin >> n;
  string l[n];
  pair<int, int> p[n];
  for (int i = 0; i < n; i++) {
    cin >> l[i];
    int x, y;
    cin >> x >> y;
    p[i] = {x, y};
  }
  int res = 0;
  for (int i = 0; i < (int)t.size(); i++) {
    for (int j = i; j < (int)t.size(); j++) {
      string par = "";
      for (int k = i; k <= j; k++) par += t[k];
      bool ok = true;
      for (int k = 0; k < n; k++) {
        s = par + "#" + l[k];
        Zfunction();
        int cont = 0;
        for (int u = 0; u < (int)s.size(); u++) {
          if (lens[u] >= (int)par.size()) cont++;
        }
        if (cont < p[k].first or cont > p[k].second) ok = false;
      }
      if (ok) ans.insert(par);
    }
  }
  cout << ans.size() << endl;
}

```