#include <bits/stdc++.h>
using namespace std;
map<pair<int, int>, long long> m[11];
map<long long, bool> M;
int main() {
  long long P[10001], l[11], r[11], ans = 0;
  string s, p[11];
  int n;
  cin >> s;
  cin >> n;
  P[0] = 1;
  for (int i = 1; i <= 10000; i++) P[i] = P[i - 1] * 63;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> l[i] >> r[i];
    for (int j = 0; j < (int)p[i].length(); j++)
      for (int k = j; k < (int)p[i].length(); k++) {
        long long H = 0;
        for (int d = j; d <= k; d++) H += p[i][d] * P[d - j];
        m[i][make_pair(j, k)] = H;
      }
  }
  for (int i = 0; i < (int)s.length(); i++) {
    for (int j = i; j < (int)s.length(); j++) {
      bool ok = true;
      long long H = 0;
      for (int d = i; d <= j; d++) {
        H += s[d] * P[d - i];
      }
      if (M[H]) continue;
      for (int d = 1; d <= n; d++) {
        int cnt = 0;
        for (int k = 0; k < (int)((int)p[d].length() - j + i); k++) {
          if (m[d][make_pair(k, k + j - i)] == H) cnt++;
        }
        if (cnt < l[d] || cnt > r[d]) {
          ok = false;
          break;
        }
      }
      if (ok) {
        ans++;
        M[H] = true;
      }
    }
  }
  cout << ans;
}

### Prompt

Create a solution in cpp for 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;
map<pair<int, int>, long long> m[11];
map<long long, bool> M;
int main() {
  long long P[10001], l[11], r[11], ans = 0;
  string s, p[11];
  int n;
  cin >> s;
  cin >> n;
  P[0] = 1;
  for (int i = 1; i <= 10000; i++) P[i] = P[i - 1] * 63;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> l[i] >> r[i];
    for (int j = 0; j < (int)p[i].length(); j++)
      for (int k = j; k < (int)p[i].length(); k++) {
        long long H = 0;
        for (int d = j; d <= k; d++) H += p[i][d] * P[d - j];
        m[i][make_pair(j, k)] = H;
      }
  }
  for (int i = 0; i < (int)s.length(); i++) {
    for (int j = i; j < (int)s.length(); j++) {
      bool ok = true;
      long long H = 0;
      for (int d = i; d <= j; d++) {
        H += s[d] * P[d - i];
      }
      if (M[H]) continue;
      for (int d = 1; d <= n; d++) {
        int cnt = 0;
        for (int k = 0; k < (int)((int)p[d].length() - j + i); k++) {
          if (m[d][make_pair(k, k + j - i)] == H) cnt++;
        }
        if (cnt < l[d] || cnt > r[d]) {
          ok = false;
          break;
        }
      }
      if (ok) {
        ans++;
        M[H] = true;
      }
    }
  }
  cout << ans;
}

```