#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-11;
const double pi = acos(-1.);
template <class T>
void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
T sqrt(T x) {
  return x * x;
}
template <class T>
T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
int countbit(int n) { return (n == 0) ? 0 : (1 + countbit(n & (n - 1))); }
int lowbit(int n) { return (n ^ (n - 1)) & n; }
typedef int Int;
Int toInt(string s) {
  istringstream sin(s);
  Int t;
  sin >> t;
  return t;
}
string toString(Int v) {
  ostringstream sout;
  sout << v;
  return sout.str();
}
template <class T>
void Out(T A[], int n) {
  for (int i = 0; i < n; ++i) cout << A[i] << ",";
  cout << endl;
}
template <class T>
void Out(vector<T> A, int n = -1) {
  if (n == -1) n = A.size();
  for (int i = 0; i < n; ++i) cout << A[i] << ",";
  cout << endl;
}
template <class T>
void Out(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
string str1[11];
int l[11], r[11];
int n;
bool isok(string &str, int k) {
  bool ret = false;
  int i, j;
  int nret = 0;
  string strtmp = str1[k];
  i = 0;
  while (i < strtmp.length()) {
    if ((i = strtmp.find(str, i)) != strtmp.npos) {
      ++nret;
      ++i;
    } else
      break;
  }
  if (nret >= l[k] && nret <= r[k]) ret = true;
  return ret;
}
int main() {
  string str;
  int i, j;
  cin >> str;
  cin >> n;
  for (i = 0; i < n; ++i) cin >> str1[i] >> l[i] >> r[i];
  map<string, int> mp;
  int ret = 0, k;
  for (i = 0; i < str.length(); ++i) {
    for (j = 1; j <= str.length() - i; ++j) {
      string sub = str.substr(i, j);
      if (!mp[sub]) {
        mp[sub] = 1;
        for (k = 0; k < n; ++k) {
          if (!isok(sub, k)) {
            break;
          }
        }
        if (k >= n) {
          ++ret;
        }
      }
    }
  }
  cout << ret << endl;
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
const double eps = 1e-11;
const double pi = acos(-1.);
template <class T>
void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
T sqrt(T x) {
  return x * x;
}
template <class T>
T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
int countbit(int n) { return (n == 0) ? 0 : (1 + countbit(n & (n - 1))); }
int lowbit(int n) { return (n ^ (n - 1)) & n; }
typedef int Int;
Int toInt(string s) {
  istringstream sin(s);
  Int t;
  sin >> t;
  return t;
}
string toString(Int v) {
  ostringstream sout;
  sout << v;
  return sout.str();
}
template <class T>
void Out(T A[], int n) {
  for (int i = 0; i < n; ++i) cout << A[i] << ",";
  cout << endl;
}
template <class T>
void Out(vector<T> A, int n = -1) {
  if (n == -1) n = A.size();
  for (int i = 0; i < n; ++i) cout << A[i] << ",";
  cout << endl;
}
template <class T>
void Out(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
string str1[11];
int l[11], r[11];
int n;
bool isok(string &str, int k) {
  bool ret = false;
  int i, j;
  int nret = 0;
  string strtmp = str1[k];
  i = 0;
  while (i < strtmp.length()) {
    if ((i = strtmp.find(str, i)) != strtmp.npos) {
      ++nret;
      ++i;
    } else
      break;
  }
  if (nret >= l[k] && nret <= r[k]) ret = true;
  return ret;
}
int main() {
  string str;
  int i, j;
  cin >> str;
  cin >> n;
  for (i = 0; i < n; ++i) cin >> str1[i] >> l[i] >> r[i];
  map<string, int> mp;
  int ret = 0, k;
  for (i = 0; i < str.length(); ++i) {
    for (j = 1; j <= str.length() - i; ++j) {
      string sub = str.substr(i, j);
      if (!mp[sub]) {
        mp[sub] = 1;
        for (k = 0; k < n; ++k) {
          if (!isok(sub, k)) {
            break;
          }
        }
        if (k >= n) {
          ++ret;
        }
      }
    }
  }
  cout << ret << endl;
  return 0;
}

```