#include <bits/stdc++.h>
using namespace std;
const long long modulo = 424242424247LL, base = 1793;
inline void doit(const string &s, unordered_map<long long, int> *mp) {
  int n = s.size();
  for (int i = 0; i < n; ++i) {
    long long h = 0;
    for (int j = i; j < n; ++j) {
      h = (h * base + s[j]) % modulo;
      ++mp[j - i + 1][h];
    }
  }
}
const int maxn = 11, maxl = 2005;
unordered_map<long long, int> met[maxn][maxl], me[maxl];
int lb[maxn], rb[maxn];
int main() {
  cin.sync_with_stdio(false);
  string s;
  cin >> s;
  doit(s, me);
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    string t;
    cin >> t >> lb[i] >> rb[i];
    doit(t, met[i]);
  }
  int ans = 0;
  for (int len = 1; len <= (int)s.size(); ++len) {
    for (auto it = me[len].begin(); it != me[len].end(); ++it) {
      long long h = it->first;
      bool ok = true;
      for (int i = 0; i < m && ok; ++i) {
        int c = met[i][len][h];
        if (c < lb[i] || c > rb[i]) ok = false;
      }
      if (ok) ++ans;
    }
  }
  cout << ans << endl;
  42;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const long long modulo = 424242424247LL, base = 1793;
inline void doit(const string &s, unordered_map<long long, int> *mp) {
  int n = s.size();
  for (int i = 0; i < n; ++i) {
    long long h = 0;
    for (int j = i; j < n; ++j) {
      h = (h * base + s[j]) % modulo;
      ++mp[j - i + 1][h];
    }
  }
}
const int maxn = 11, maxl = 2005;
unordered_map<long long, int> met[maxn][maxl], me[maxl];
int lb[maxn], rb[maxn];
int main() {
  cin.sync_with_stdio(false);
  string s;
  cin >> s;
  doit(s, me);
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    string t;
    cin >> t >> lb[i] >> rb[i];
    doit(t, met[i]);
  }
  int ans = 0;
  for (int len = 1; len <= (int)s.size(); ++len) {
    for (auto it = me[len].begin(); it != me[len].end(); ++it) {
      long long h = it->first;
      bool ok = true;
      for (int i = 0; i < m && ok; ++i) {
        int c = met[i][len][h];
        if (c < lb[i] || c > rb[i]) ok = false;
      }
      if (ok) ++ans;
    }
  }
  cout << ans << endl;
  42;
  return 0;
}

```