#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
int count(string s1, string s2) {
  string s = s1 + "#";
  vector<int> pref(s1.length() + 1, 0);
  int lp = 0;
  for (int i = 1; i < s.length(); i++) {
    while (lp > 0 && s[i] != s[lp]) lp = pref[lp - 1];
    if (s[i] == s[lp]) lp++;
    pref[i] = lp;
  }
  lp = 0;
  int res = 0;
  for (int i = 0; i < s2.length(); i++) {
    while (lp > 0 && s2[i] != s[lp]) lp = pref[lp - 1];
    if (s2[i] == s[lp]) lp++;
    if (lp == s1.length()) res++;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  set<string> was;
  int n;
  cin >> n;
  vector<string> p(n);
  vector<int> l(n), r(n);
  for (int i = 0; i < n; i++) cin >> p[i] >> l[i] >> r[i];
  string s2;
  for (int i = 0; i < s.length(); i++) {
    s2 = "";
    for (int j = i; j < s.length(); j++) {
      s2 += s[j];
      was.insert(s2);
    }
  }
  set<string>::iterator it = was.begin();
  int cnt = 0, q, w;
  while (it != was.end()) {
    bool fin = true;
    for (int i = 0; i < n; i++) {
      w = count(*it, p[i]);
      if (w < l[i] || w > r[i]) {
        fin = false;
        break;
      }
    }
    if (fin) cnt++;
    it++;
  }
  cout << cnt;
}

### Prompt

In Cpp, your task is to solve 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:64000000")
using namespace std;
int count(string s1, string s2) {
  string s = s1 + "#";
  vector<int> pref(s1.length() + 1, 0);
  int lp = 0;
  for (int i = 1; i < s.length(); i++) {
    while (lp > 0 && s[i] != s[lp]) lp = pref[lp - 1];
    if (s[i] == s[lp]) lp++;
    pref[i] = lp;
  }
  lp = 0;
  int res = 0;
  for (int i = 0; i < s2.length(); i++) {
    while (lp > 0 && s2[i] != s[lp]) lp = pref[lp - 1];
    if (s2[i] == s[lp]) lp++;
    if (lp == s1.length()) res++;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  set<string> was;
  int n;
  cin >> n;
  vector<string> p(n);
  vector<int> l(n), r(n);
  for (int i = 0; i < n; i++) cin >> p[i] >> l[i] >> r[i];
  string s2;
  for (int i = 0; i < s.length(); i++) {
    s2 = "";
    for (int j = i; j < s.length(); j++) {
      s2 += s[j];
      was.insert(s2);
    }
  }
  set<string>::iterator it = was.begin();
  int cnt = 0, q, w;
  while (it != was.end()) {
    bool fin = true;
    for (int i = 0; i < n; i++) {
      w = count(*it, p[i]);
      if (w < l[i] || w > r[i]) {
        fin = false;
        break;
      }
    }
    if (fin) cnt++;
    it++;
  }
  cout << cnt;
}

```