#include <bits/stdc++.h>
using namespace std;
struct node {
  string str;
  int l;
  int r;
} a[15];
int main() {
  string s;
  int n;
  cin >> s;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i].str >> a[i].l >> a[i].r;
  int ans = 0;
  map<string, int> mp;
  for (int i = 0; i < s.size(); i++)
    for (int j = 1; j <= s.size() - i; j++) {
      string str = s.substr(i, j);
      int k;
      for (k = 0; k < n; k++) {
        int cnt = 0;
        int tmp;
        for (int v = 0; v < a[k].str.size();) {
          tmp = a[k].str.find(str, v);
          if (tmp == string::npos)
            break;
          else {
            v = tmp + 1;
            cnt++;
          }
        }
        if (cnt < a[k].l || cnt > a[k].r) break;
      }
      if (k == n) {
        mp[str] = 1;
      }
    }
  cout << mp.size() << 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;
struct node {
  string str;
  int l;
  int r;
} a[15];
int main() {
  string s;
  int n;
  cin >> s;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i].str >> a[i].l >> a[i].r;
  int ans = 0;
  map<string, int> mp;
  for (int i = 0; i < s.size(); i++)
    for (int j = 1; j <= s.size() - i; j++) {
      string str = s.substr(i, j);
      int k;
      for (k = 0; k < n; k++) {
        int cnt = 0;
        int tmp;
        for (int v = 0; v < a[k].str.size();) {
          tmp = a[k].str.find(str, v);
          if (tmp == string::npos)
            break;
          else {
            v = tmp + 1;
            cnt++;
          }
        }
        if (cnt < a[k].l || cnt > a[k].r) break;
      }
      if (k == n) {
        mp[str] = 1;
      }
    }
  cout << mp.size() << endl;
  return 0;
}

```