#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:65000000")
using namespace std;
vector<int> z_function(const string &s) {
  int n = (int)s.length();
  vector<int> z(n);
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
int test(const string &s, int l) {
  vector<int> z = z_function(s);
  int answ = 0;
  for (int i = 0; i < z.size(); ++i)
    if (z[i] == l) ++answ;
  return answ;
}
vector<string> vs;
vector<pair<int, int> > vp;
int n;
string s;
set<string> st;
int main() {
  cin >> s;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    string t;
    pair<int, int> p;
    cin >> t >> p.first >> p.second;
    vs.push_back(t);
    vp.push_back(p);
  }
  int answ = 0;
  for (int i = 0; i < s.size(); ++i)
    for (int j = i + 1; j <= s.size(); ++j) {
      string t = s.substr(i, j - i);
      st.insert(t);
    }
  for (set<string>::iterator it = st.begin(); it != st.end(); ++it) {
    string t = *it;
    bool b = 1;
    for (int k = 0; b && k < n; ++k) {
      string r = t + "#" + vs[k];
      int a = test(r, t.length());
      if (a < vp[k].first || a > vp[k].second) b = false;
    }
    if (b) ++answ;
  }
  cout << answ << endl;
  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>
#pragma comment(linker, "/STACK:65000000")
using namespace std;
vector<int> z_function(const string &s) {
  int n = (int)s.length();
  vector<int> z(n);
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
int test(const string &s, int l) {
  vector<int> z = z_function(s);
  int answ = 0;
  for (int i = 0; i < z.size(); ++i)
    if (z[i] == l) ++answ;
  return answ;
}
vector<string> vs;
vector<pair<int, int> > vp;
int n;
string s;
set<string> st;
int main() {
  cin >> s;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    string t;
    pair<int, int> p;
    cin >> t >> p.first >> p.second;
    vs.push_back(t);
    vp.push_back(p);
  }
  int answ = 0;
  for (int i = 0; i < s.size(); ++i)
    for (int j = i + 1; j <= s.size(); ++j) {
      string t = s.substr(i, j - i);
      st.insert(t);
    }
  for (set<string>::iterator it = st.begin(); it != st.end(); ++it) {
    string t = *it;
    bool b = 1;
    for (int k = 0; b && k < n; ++k) {
      string r = t + "#" + vs[k];
      int a = test(r, t.length());
      if (a < vp[k].first || a > vp[k].second) b = false;
    }
    if (b) ++answ;
  }
  cout << answ << endl;
  return 0;
}

```