#include <bits/stdc++.h>
using namespace std;
const long long MX = 100000;
const long long MOD = 1000000007;
const long long base = 257;
long long fast_exp(long long base, long long exp, long long mod) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % mod;
    base = (base * base) % mod;
    exp /= 2;
  }
  return res;
}
long long val(string s) {
  int n = (int)s.size();
  long long ans = 0;
  for (int i = 0, _i = (n); i < _i; ++i) {
    ans = (ans * base + (s[i] - 'a' + 1)) % MOD;
  }
  return ans;
}
bool Rabin(string s, string p, long long l, long long r) {
  long long check = val(s);
  long long ini = 0, cnt = 0, k = (int)s.size(), n = (int)p.size(), i = 0;
  if ((int)p.size() < (int)s.size()) return false;
  for (i = 0; i < k; i++) ini = (ini * base + (p[i] - 'a' + 1)) % MOD;
  if (ini == check) ++cnt;
  for (; i < n && (i - k >= 0); i++) {
    ini = (ini - (p[i - k] - 'a' + 1) * (fast_exp(base, k, MOD)));
    while (ini >= MOD) ini -= MOD;
    ini = (ini * base + (p[i] - 'a' + 1));
    if (ini == check) ++cnt;
  }
  return (cnt >= l && cnt <= r);
}
vector<string> v(12);
vector<pair<long long, long long> > in(12);
map<long long, long long> arr[15];
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin.tie(NULL);
  string s, p;
  long long l, r, n;
  cin >> s;
  cin >> n;
  int N = (int)s.size();
  set<long long> hash;
  for (int i = 0; i < N; i++) {
    long long ans = 0;
    for (int j = i; j < N; j++) {
      ans = (ans * base + (s[j] - 'a' + 1)) % MOD;
      hash.insert(ans);
    }
  }
  for (int i = 0, _i = (n); i < _i; ++i) {
    cin >> v[i] >> in[i].first >> in[i].second;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (int)v[i].size(); j++) {
      long long ans = 0;
      for (int k = j; k < (int)v[i].size(); k++) {
        ans = (ans * base + (v[i][k] - 'a' + 1)) % MOD;
        ++arr[i][ans];
      }
    }
  }
  long long cnt = 0;
  for (auto it = hash.begin(); it != hash.end(); it++) {
    bool flag = true;
    for (int i = 0; i < n; i++) {
      flag =
          flag && (arr[i][*it] >= in[i].first && arr[i][*it] <= in[i].second);
    }
    if (flag) ++cnt;
  }
  cout << cnt;
  return 0;
}

### Prompt

Please create a solution in cpp 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 long long MX = 100000;
const long long MOD = 1000000007;
const long long base = 257;
long long fast_exp(long long base, long long exp, long long mod) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % mod;
    base = (base * base) % mod;
    exp /= 2;
  }
  return res;
}
long long val(string s) {
  int n = (int)s.size();
  long long ans = 0;
  for (int i = 0, _i = (n); i < _i; ++i) {
    ans = (ans * base + (s[i] - 'a' + 1)) % MOD;
  }
  return ans;
}
bool Rabin(string s, string p, long long l, long long r) {
  long long check = val(s);
  long long ini = 0, cnt = 0, k = (int)s.size(), n = (int)p.size(), i = 0;
  if ((int)p.size() < (int)s.size()) return false;
  for (i = 0; i < k; i++) ini = (ini * base + (p[i] - 'a' + 1)) % MOD;
  if (ini == check) ++cnt;
  for (; i < n && (i - k >= 0); i++) {
    ini = (ini - (p[i - k] - 'a' + 1) * (fast_exp(base, k, MOD)));
    while (ini >= MOD) ini -= MOD;
    ini = (ini * base + (p[i] - 'a' + 1));
    if (ini == check) ++cnt;
  }
  return (cnt >= l && cnt <= r);
}
vector<string> v(12);
vector<pair<long long, long long> > in(12);
map<long long, long long> arr[15];
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin.tie(NULL);
  string s, p;
  long long l, r, n;
  cin >> s;
  cin >> n;
  int N = (int)s.size();
  set<long long> hash;
  for (int i = 0; i < N; i++) {
    long long ans = 0;
    for (int j = i; j < N; j++) {
      ans = (ans * base + (s[j] - 'a' + 1)) % MOD;
      hash.insert(ans);
    }
  }
  for (int i = 0, _i = (n); i < _i; ++i) {
    cin >> v[i] >> in[i].first >> in[i].second;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (int)v[i].size(); j++) {
      long long ans = 0;
      for (int k = j; k < (int)v[i].size(); k++) {
        ans = (ans * base + (v[i][k] - 'a' + 1)) % MOD;
        ++arr[i][ans];
      }
    }
  }
  long long cnt = 0;
  for (auto it = hash.begin(); it != hash.end(); it++) {
    bool flag = true;
    for (int i = 0; i < n; i++) {
      flag =
          flag && (arr[i][*it] >= in[i].first && arr[i][*it] <= in[i].second);
    }
    if (flag) ++cnt;
  }
  cout << cnt;
  return 0;
}

```