#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
const int INF = (int)1e9;
const long long INF64 = (long long)1e18;
const long double eps = 1e-9;
const long double pi = 3.14159265358979323846;
int n;
string s;
string g[20];
int l[20], r[20];
vector<int> arr[20];
vector<int> sums[20];
vector<int> prefix_function(string s) {
  int n = (int)s.length();
  vector<int> pi(n);
  for (int i = 1; i < n; ++i) {
    int j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) j = pi[j - 1];
    if (s[i] == s[j]) ++j;
    pi[i] = j;
  }
  return pi;
}
vector<int> build(int pos) {
  vector<int> pf = prefix_function(g[pos] + "#" + s);
  vector<int> res;
  for (int i = 0; i < int((int)(s).size()); ++i) {
    int ppos = (int)(g[pos]).size() + i + 1;
    if (pf[ppos] == (int)(g[pos]).size())
      res.push_back(1);
    else
      res.push_back(0);
  }
  return res;
}
int main() {
  cin >> s >> n;
  for (int i = 0; i < int(n); ++i) {
    cin >> g[i] >> l[i] >> r[i];
  }
  set<string> answer;
  for (int i = 0; i < int((int)(s).size()); ++i) {
    string t = "";
    for (int j = i; j < (int)(s).size(); j++) {
      t += s[j];
      bool ok = true;
      for (int k = 0; k < int(n); ++k) {
        vector<int> pf = prefix_function(t + "#" + g[k]);
        int cnt = 0;
        for (int r = 0; r < int((int)(pf).size()); ++r) {
          if (pf[r] == (int)(t).size()) cnt++;
        }
        if (!(cnt >= l[k] && cnt <= r[k])) ok = false;
      }
      if (ok) answer.insert(t);
    }
  }
  cout << (int)(answer).size();
}

### Prompt

Your task is to create 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>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
const int INF = (int)1e9;
const long long INF64 = (long long)1e18;
const long double eps = 1e-9;
const long double pi = 3.14159265358979323846;
int n;
string s;
string g[20];
int l[20], r[20];
vector<int> arr[20];
vector<int> sums[20];
vector<int> prefix_function(string s) {
  int n = (int)s.length();
  vector<int> pi(n);
  for (int i = 1; i < n; ++i) {
    int j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) j = pi[j - 1];
    if (s[i] == s[j]) ++j;
    pi[i] = j;
  }
  return pi;
}
vector<int> build(int pos) {
  vector<int> pf = prefix_function(g[pos] + "#" + s);
  vector<int> res;
  for (int i = 0; i < int((int)(s).size()); ++i) {
    int ppos = (int)(g[pos]).size() + i + 1;
    if (pf[ppos] == (int)(g[pos]).size())
      res.push_back(1);
    else
      res.push_back(0);
  }
  return res;
}
int main() {
  cin >> s >> n;
  for (int i = 0; i < int(n); ++i) {
    cin >> g[i] >> l[i] >> r[i];
  }
  set<string> answer;
  for (int i = 0; i < int((int)(s).size()); ++i) {
    string t = "";
    for (int j = i; j < (int)(s).size(); j++) {
      t += s[j];
      bool ok = true;
      for (int k = 0; k < int(n); ++k) {
        vector<int> pf = prefix_function(t + "#" + g[k]);
        int cnt = 0;
        for (int r = 0; r < int((int)(pf).size()); ++r) {
          if (pf[r] == (int)(t).size()) cnt++;
        }
        if (!(cnt >= l[k] && cnt <= r[k])) ok = false;
      }
      if (ok) answer.insert(t);
    }
  }
  cout << (int)(answer).size();
}

```