#include <bits/stdc++.h>
using namespace std;
int T[1000];
void computeT(string &Needle) {
  int i = 0, j = -1;
  T[i] = j;
  while (i < Needle.size()) {
    while (j >= 0 && Needle[i] != Needle[j]) j = T[j];
    T[++i] = ++j;
  }
}
int findOcurrence(string &Needle, string &Hay) {
  int i = 0, j = 0, r = 0;
  computeT(Needle);
  while (i < Hay.size()) {
    while (j >= 0 && Hay[i] != Needle[j]) j = T[j];
    i++;
    j++;
    if (j == Needle.size()) {
      r++;
      j = T[j];
    }
  }
  return r;
}
string P[20];
int L[20], R[20];
int main() {
  string A;
  cin >> A;
  set<string> S;
  for (int i = 0; i < A.size(); ++i)
    for (int j = i; j < A.size(); ++j) S.insert(A.substr(i, j - i + 1));
  int N;
  cin >> N;
  for (int i = 0; i < N; ++i) cin >> P[i] >> L[i] >> R[i];
  int s = 0;
  for (set<string>::iterator it = S.begin(); it != S.end(); ++it) {
    int ok = 1;
    for (int i = 0; i < N && ok; ++i) {
      string a = *it;
      int t = findOcurrence(a, P[i]);
      if (t < L[i] || t > R[i]) ok = 0;
    }
    s += ok;
  }
  cout << s << endl;
}

### Prompt

Create a solution in cpp for 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 T[1000];
void computeT(string &Needle) {
  int i = 0, j = -1;
  T[i] = j;
  while (i < Needle.size()) {
    while (j >= 0 && Needle[i] != Needle[j]) j = T[j];
    T[++i] = ++j;
  }
}
int findOcurrence(string &Needle, string &Hay) {
  int i = 0, j = 0, r = 0;
  computeT(Needle);
  while (i < Hay.size()) {
    while (j >= 0 && Hay[i] != Needle[j]) j = T[j];
    i++;
    j++;
    if (j == Needle.size()) {
      r++;
      j = T[j];
    }
  }
  return r;
}
string P[20];
int L[20], R[20];
int main() {
  string A;
  cin >> A;
  set<string> S;
  for (int i = 0; i < A.size(); ++i)
    for (int j = i; j < A.size(); ++j) S.insert(A.substr(i, j - i + 1));
  int N;
  cin >> N;
  for (int i = 0; i < N; ++i) cin >> P[i] >> L[i] >> R[i];
  int s = 0;
  for (set<string>::iterator it = S.begin(); it != S.end(); ++it) {
    int ok = 1;
    for (int i = 0; i < N && ok; ++i) {
      string a = *it;
      int t = findOcurrence(a, P[i]);
      if (t < L[i] || t > R[i]) ok = 0;
    }
    s += ok;
  }
  cout << s << endl;
}

```