#include <bits/stdc++.h>
using namespace std;
string s, p[1111], t;
int l[111], r[111], n, pp[1111], res;
map<string, bool> m;
int main() {
  cin >> s;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i] >> l[i] >> r[i];
  for (int i = 0; i < ((int)(s).size()); i++) {
    t = "";
    for (int j = i; j < ((int)(s).size()); j++) {
      t += s[j];
      int y = 0;
      if (m.find(t) == m.end()) {
        m[t] = 1;
        for (int h = 1; h <= n; h++) {
          string tt = t;
          tt += '#';
          tt += p[h];
          for (int u = 0; u <= ((int)(tt).size()); u++) pp[u] = 0;
          for (int u = 1; u < ((int)(tt).size()); u++) {
            int k = pp[u - 1];
            while (k > 0 && tt[k] != tt[u]) k = pp[k - 1];
            if (tt[u] == tt[k]) k++;
            pp[u] = k;
          }
          int o = 0;
          for (int u = 0; u < ((int)(tt).size()); u++)
            if (pp[u] == ((int)(t).size())) o++;
          if (l[h] <= o && o <= r[h]) y++;
        }
        if (y == n) {
          res++;
        }
      }
    }
  }
  cout << res;
  return 0;
}

### Prompt

Your task is to create 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;
string s, p[1111], t;
int l[111], r[111], n, pp[1111], res;
map<string, bool> m;
int main() {
  cin >> s;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i] >> l[i] >> r[i];
  for (int i = 0; i < ((int)(s).size()); i++) {
    t = "";
    for (int j = i; j < ((int)(s).size()); j++) {
      t += s[j];
      int y = 0;
      if (m.find(t) == m.end()) {
        m[t] = 1;
        for (int h = 1; h <= n; h++) {
          string tt = t;
          tt += '#';
          tt += p[h];
          for (int u = 0; u <= ((int)(tt).size()); u++) pp[u] = 0;
          for (int u = 1; u < ((int)(tt).size()); u++) {
            int k = pp[u - 1];
            while (k > 0 && tt[k] != tt[u]) k = pp[k - 1];
            if (tt[u] == tt[k]) k++;
            pp[u] = k;
          }
          int o = 0;
          for (int u = 0; u < ((int)(tt).size()); u++)
            if (pp[u] == ((int)(t).size())) o++;
          if (l[h] <= o && o <= r[h]) y++;
        }
        if (y == n) {
          res++;
        }
      }
    }
  }
  cout << res;
  return 0;
}

```