#include <bits/stdc++.h>
using namespace std;
vector<int> fail_fun(string pt) {
  int m = ((int)(pt.size())), mat;
  vector<int> f(m + 1, -1);
  for (int i(1), _b(m + 1); i < _b; ++i) {
    mat = f[i - 1];
    while (mat != -1 && pt[mat] != pt[i - 1]) mat = f[mat];
    f[i] = mat + 1;
  }
  return f;
}
vector<int> kmp(string str, string ptrn) {
  vector<int> ret;
  int str_pos = 0, ptrn_pos = 0, ptrn_sz = ((int)(ptrn.size()));
  vector<int> f = fail_fun(ptrn);
  for (int i(0), _n(((int)(str.size()))); i < _n; ++i) {
    while (ptrn_pos != -1 &&
           (ptrn_pos == ptrn_sz || ptrn[ptrn_pos] != str[str_pos]))
      ptrn_pos = f[ptrn_pos];
    str_pos++;
    ptrn_pos++;
    if (ptrn_pos == ptrn_sz) ret.push_back(str_pos - ptrn_sz);
  }
  return ret;
}
set<string> st;
string s, p[11];
int l[11], r[11];
int n;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  cin >> n;
  for (int i(0), _n(n); i < _n; ++i) {
    cin >> p[i] >> l[i] >> r[i];
  }
  for (int i(0), _n(((int)(s.size()))); i < _n; ++i) {
    for (int j(1), _b(((int)(s.size())) + 1); j < _b; ++j) {
      if (i + j <= ((int)(s.size()))) st.insert(s.substr(i, j));
    }
  }
  int ans = 0;
  for (__typeof((st).begin()) i = (st).begin(); i != (st).end(); ++i) {
    int f = 1, t1;
    for (int j(0), _n(n); j < _n; ++j) {
      t1 = ((int)(kmp(p[j], *i).size()));
      if (t1 < l[j] || t1 > r[j]) f = 0;
    }
    ans += f;
  }
  cout << ans;
}

### Prompt

Develop a solution in CPP 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;
vector<int> fail_fun(string pt) {
  int m = ((int)(pt.size())), mat;
  vector<int> f(m + 1, -1);
  for (int i(1), _b(m + 1); i < _b; ++i) {
    mat = f[i - 1];
    while (mat != -1 && pt[mat] != pt[i - 1]) mat = f[mat];
    f[i] = mat + 1;
  }
  return f;
}
vector<int> kmp(string str, string ptrn) {
  vector<int> ret;
  int str_pos = 0, ptrn_pos = 0, ptrn_sz = ((int)(ptrn.size()));
  vector<int> f = fail_fun(ptrn);
  for (int i(0), _n(((int)(str.size()))); i < _n; ++i) {
    while (ptrn_pos != -1 &&
           (ptrn_pos == ptrn_sz || ptrn[ptrn_pos] != str[str_pos]))
      ptrn_pos = f[ptrn_pos];
    str_pos++;
    ptrn_pos++;
    if (ptrn_pos == ptrn_sz) ret.push_back(str_pos - ptrn_sz);
  }
  return ret;
}
set<string> st;
string s, p[11];
int l[11], r[11];
int n;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  cin >> n;
  for (int i(0), _n(n); i < _n; ++i) {
    cin >> p[i] >> l[i] >> r[i];
  }
  for (int i(0), _n(((int)(s.size()))); i < _n; ++i) {
    for (int j(1), _b(((int)(s.size())) + 1); j < _b; ++j) {
      if (i + j <= ((int)(s.size()))) st.insert(s.substr(i, j));
    }
  }
  int ans = 0;
  for (__typeof((st).begin()) i = (st).begin(); i != (st).end(); ++i) {
    int f = 1, t1;
    for (int j(0), _n(n); j < _n; ++j) {
      t1 = ((int)(kmp(p[j], *i).size()));
      if (t1 < l[j] || t1 > r[j]) f = 0;
    }
    ans += f;
  }
  cout << ans;
}

```