#include <bits/stdc++.h>
using namespace std;
int count(const string& s, const string& p) {
  string t = ' ' + p + '#' + s;
  vector<int> k(t.size());
  int res = 0;
  for (int i = 2; i < t.size(); ++i) {
    int x = k[i - 1];
    while (t[x + 1] != t[i] && x > 0) x = k[x];
    if (t[x + 1] == t[i]) k[i] = x + 1;
    if (k[i] == p.length()) res++;
  }
  return res;
}
void solve() {
  string s;
  int n;
  cin >> s;
  cin >> n;
  vector<string> p(n);
  vector<int> l(n);
  vector<int> r(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i] >> l[i] >> r[i];
  }
  int ans = 0;
  set<long long> st;
  const int K = 100;
  for (int o = 0; o < s.length(); ++o) {
    int L = 1;
    int R = s.length() - o;
    for (int i = 0; i < n; ++i) {
      while (L + K <= R && count(p[i], s.substr(o, L + K - 1)) > r[i]) L += K;
      while (L <= R && count(p[i], s.substr(o, L)) > r[i]) L++;
      while (L <= R - K && count(p[i], s.substr(o, R - K + 1)) < l[i]) R -= K;
      while (L <= R && count(p[i], s.substr(o, R)) < l[i]) R--;
    }
    ans += R - L + 1;
    long long HS = 0;
    for (int i = 0; i < R; ++i) {
      HS = HS * 299 + s[i + o];
      if (i + 1 >= L) st.insert(HS);
    }
  }
  cout << st.size() << endl;
}
int main() {
  solve();
  return 0;
}

### 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 count(const string& s, const string& p) {
  string t = ' ' + p + '#' + s;
  vector<int> k(t.size());
  int res = 0;
  for (int i = 2; i < t.size(); ++i) {
    int x = k[i - 1];
    while (t[x + 1] != t[i] && x > 0) x = k[x];
    if (t[x + 1] == t[i]) k[i] = x + 1;
    if (k[i] == p.length()) res++;
  }
  return res;
}
void solve() {
  string s;
  int n;
  cin >> s;
  cin >> n;
  vector<string> p(n);
  vector<int> l(n);
  vector<int> r(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i] >> l[i] >> r[i];
  }
  int ans = 0;
  set<long long> st;
  const int K = 100;
  for (int o = 0; o < s.length(); ++o) {
    int L = 1;
    int R = s.length() - o;
    for (int i = 0; i < n; ++i) {
      while (L + K <= R && count(p[i], s.substr(o, L + K - 1)) > r[i]) L += K;
      while (L <= R && count(p[i], s.substr(o, L)) > r[i]) L++;
      while (L <= R - K && count(p[i], s.substr(o, R - K + 1)) < l[i]) R -= K;
      while (L <= R && count(p[i], s.substr(o, R)) < l[i]) R--;
    }
    ans += R - L + 1;
    long long HS = 0;
    for (int i = 0; i < R; ++i) {
      HS = HS * 299 + s[i + o];
      if (i + 1 >= L) st.insert(HS);
    }
  }
  cout << st.size() << endl;
}
int main() {
  solve();
  return 0;
}

```