#include <bits/stdc++.h>
using namespace std;
long long GCD(long long a, long long b) { return b != 0 ? GCD(b, a % b) : a; }
long long LCM(long long a, long long b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
int back[205];
void KMPInit(string& s) {
  int i = 0, j = -1;
  back[0] = -1;
  while (i < (int)(s).size()) {
    while (j >= 0 && s[i] != s[j]) j = back[j];
    ++i;
    ++j;
    back[i] = j;
  }
}
int KMP(string& needle, string& hey) {
  int i = 0, j = 0, r = 0;
  while (i < (int)(hey).size()) {
    while (j >= 0 && hey[i] != needle[j]) j = back[j];
    ++i;
    ++j;
    if (j == (int)(needle).size()) {
      ++r;
      j = back[j];
    }
  }
  return r;
}
string p[11];
int l[11], r[11];
int main() {
  string s;
  getline(cin, s);
  int N;
  scanf("%d", &N);
  for (int i = 0, max_ = (N); i < max_; ++i) cin >> p[i] >> l[i] >> r[i];
  set<string> m;
  int res = 0;
  for (int i = 0, max_ = ((int)(s).size()); i < max_; ++i)
    for (int j = (i), max_ = ((int)(s).size()); j < max_; ++j) {
      string str = s.substr(i, j - i + 1);
      KMPInit(str);
      bool ok = true;
      for (int k = 0, max_ = (N); k < max_; ++k) {
        int occur = KMP(str, p[k]);
        if (occur < l[k] || occur > r[k]) {
          ok = false;
          break;
        }
      }
      if (ok) m.insert(str);
    }
  printf("%d\n", (int)(m).size());
  return 0;
}

### Prompt

Please create a solution in cpp 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;
long long GCD(long long a, long long b) { return b != 0 ? GCD(b, a % b) : a; }
long long LCM(long long a, long long b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
int back[205];
void KMPInit(string& s) {
  int i = 0, j = -1;
  back[0] = -1;
  while (i < (int)(s).size()) {
    while (j >= 0 && s[i] != s[j]) j = back[j];
    ++i;
    ++j;
    back[i] = j;
  }
}
int KMP(string& needle, string& hey) {
  int i = 0, j = 0, r = 0;
  while (i < (int)(hey).size()) {
    while (j >= 0 && hey[i] != needle[j]) j = back[j];
    ++i;
    ++j;
    if (j == (int)(needle).size()) {
      ++r;
      j = back[j];
    }
  }
  return r;
}
string p[11];
int l[11], r[11];
int main() {
  string s;
  getline(cin, s);
  int N;
  scanf("%d", &N);
  for (int i = 0, max_ = (N); i < max_; ++i) cin >> p[i] >> l[i] >> r[i];
  set<string> m;
  int res = 0;
  for (int i = 0, max_ = ((int)(s).size()); i < max_; ++i)
    for (int j = (i), max_ = ((int)(s).size()); j < max_; ++j) {
      string str = s.substr(i, j - i + 1);
      KMPInit(str);
      bool ok = true;
      for (int k = 0, max_ = (N); k < max_; ++k) {
        int occur = KMP(str, p[k]);
        if (occur < l[k] || occur > r[k]) {
          ok = false;
          break;
        }
      }
      if (ok) m.insert(str);
    }
  printf("%d\n", (int)(m).size());
  return 0;
}

```