#include <bits/stdc++.h>
using namespace std;
template <class A, class B>
ostream& operator<<(ostream& o, pair<A, B>& p) {
  return o << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "{";
  for (typeof((v).begin()) it = (v).begin(); it != (v).end(); it++)
    o << *it << ", ";
  return o << "}";
}
struct debugger {
  ostream& out;
  bool first;
  void start() { first = true; }
  debugger(ostream& out) : out(out) { start(); }
  void nl() {
    out << endl;
    start();
  }
  template <class T>
  debugger& operator,(T& v) {
    if (first)
      first = false;
    else
      out << ", ";
    out << v;
    return *this;
  }
  template <class T>
  debugger& operator,(pair<T*, int> arr) {
    out << "{";
    for (typeof((arr.second) - 1) i = (0); i <= ((arr.second) - 1); i++) {
      out << arr.first[i] << ", ";
    }
    out << "}";
    return *this;
  }
} dbg(cerr);
int phi[222];
int count(string& text, string& pat) {
  int i, k, n = (int)pat.size(), c = 0;
  phi[0] = phi[1] = k = 0;
  for (i = 1; i < (int)pat.size(); i++) {
    while (k > 0 && pat[k] != pat[i]) k = phi[k];
    phi[i + 1] = pat[k] == pat[i] ? ++k : 0;
  }
  for (i = k = 0; i < (int)text.size(); i++) {
    while (k > 0 && pat[k] != text[i]) k = phi[k];
    if (pat[k] == text[i]) ++k;
    if (k == (int)pat.size()) {
      c++;
      k = phi[k];
    }
  }
  return c;
}
int main() {
  string s, p[12];
  int n, l[12], r[12];
  cin >> s;
  cin >> n;
  for (typeof((n)-1) i = (0); i <= ((n)-1); i++) {
    cin >> p[i] >> l[i] >> r[i];
  }
  int ans = 0;
  map<string, int> mp;
  for (typeof(((int)s.size()) - 1) i = (0); i <= (((int)s.size()) - 1); i++) {
    for (typeof((i + 1) - 1) j = (0); j <= ((i + 1) - 1); j++) {
      string st = s.substr(j, i - j + 1);
      if (mp[st] == 0) {
        mp[st] = 1;
        bool ok = true;
        for (typeof((n)-1) k = (0); k <= ((n)-1); k++) {
          int x = count(p[k], st);
          if (!((l[k]) <= (x) && (x) <= (r[k]))) ok = false;
        }
        if (ok) {
          ans++;
        }
      }
    }
  }
  cout << ans << endl;
  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;
template <class A, class B>
ostream& operator<<(ostream& o, pair<A, B>& p) {
  return o << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "{";
  for (typeof((v).begin()) it = (v).begin(); it != (v).end(); it++)
    o << *it << ", ";
  return o << "}";
}
struct debugger {
  ostream& out;
  bool first;
  void start() { first = true; }
  debugger(ostream& out) : out(out) { start(); }
  void nl() {
    out << endl;
    start();
  }
  template <class T>
  debugger& operator,(T& v) {
    if (first)
      first = false;
    else
      out << ", ";
    out << v;
    return *this;
  }
  template <class T>
  debugger& operator,(pair<T*, int> arr) {
    out << "{";
    for (typeof((arr.second) - 1) i = (0); i <= ((arr.second) - 1); i++) {
      out << arr.first[i] << ", ";
    }
    out << "}";
    return *this;
  }
} dbg(cerr);
int phi[222];
int count(string& text, string& pat) {
  int i, k, n = (int)pat.size(), c = 0;
  phi[0] = phi[1] = k = 0;
  for (i = 1; i < (int)pat.size(); i++) {
    while (k > 0 && pat[k] != pat[i]) k = phi[k];
    phi[i + 1] = pat[k] == pat[i] ? ++k : 0;
  }
  for (i = k = 0; i < (int)text.size(); i++) {
    while (k > 0 && pat[k] != text[i]) k = phi[k];
    if (pat[k] == text[i]) ++k;
    if (k == (int)pat.size()) {
      c++;
      k = phi[k];
    }
  }
  return c;
}
int main() {
  string s, p[12];
  int n, l[12], r[12];
  cin >> s;
  cin >> n;
  for (typeof((n)-1) i = (0); i <= ((n)-1); i++) {
    cin >> p[i] >> l[i] >> r[i];
  }
  int ans = 0;
  map<string, int> mp;
  for (typeof(((int)s.size()) - 1) i = (0); i <= (((int)s.size()) - 1); i++) {
    for (typeof((i + 1) - 1) j = (0); j <= ((i + 1) - 1); j++) {
      string st = s.substr(j, i - j + 1);
      if (mp[st] == 0) {
        mp[st] = 1;
        bool ok = true;
        for (typeof((n)-1) k = (0); k <= ((n)-1); k++) {
          int x = count(p[k], st);
          if (!((l[k]) <= (x) && (x) <= (r[k]))) ok = false;
        }
        if (ok) {
          ans++;
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```