#include <bits/stdc++.h>
using namespace std;
char A[202], C[202];
int N;
char B[12][202];
int L[12], R[12];
int prefix[202], P[12][202];
int result;
set<string> M;
string SNOW;
int main() {
  cin >> (A + 1) >> N;
  for (int i = 1; i <= N; ++i) {
    cin >> (B[i] + 1) >> L[i] >> R[i];
    int pnow = 0;
    for (int j = 2; B[i][j] != 0; ++j) {
      while (pnow != 0 && B[i][pnow + 1] != B[i][j]) pnow = P[i][pnow];
      if (B[i][pnow + 1] == B[i][j]) ++pnow;
      P[i][j] = pnow;
    }
  }
  for (int i = 1; A[i] != 0; ++i)
    for (int j = i; A[j] != 0; ++j) {
      SNOW.clear();
      for (int k = i; k <= j; ++k) {
        C[k - i + 1] = A[k];
        SNOW += A[k];
      }
      memset(prefix, 0, sizeof(prefix));
      int pnow = 0;
      for (int k = 2; k <= j - i + 1; ++k) {
        while (pnow != 0 && C[pnow + 1] != C[k]) pnow = prefix[pnow];
        if (C[pnow + 1] == C[k]) ++pnow;
        prefix[k] = pnow;
      }
      bool ok = true;
      for (int k = 1; k <= N; ++k) {
        int matchnow = 0, matches = 0;
        for (int l = 1; B[k][l] != 0; ++l) {
          while (matchnow != 0 && C[matchnow + 1] != B[k][l])
            matchnow = prefix[matchnow];
          if (C[matchnow + 1] == B[k][l]) ++matchnow;
          if (matchnow == j - i + 1) {
            ++matches;
            matchnow = prefix[matchnow];
          }
        }
        if (matches < L[k] || matches > R[k]) {
          ok = false;
          break;
        }
      }
      if (ok) M.insert(SNOW);
    }
  cout << M.size();
}

### 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;
char A[202], C[202];
int N;
char B[12][202];
int L[12], R[12];
int prefix[202], P[12][202];
int result;
set<string> M;
string SNOW;
int main() {
  cin >> (A + 1) >> N;
  for (int i = 1; i <= N; ++i) {
    cin >> (B[i] + 1) >> L[i] >> R[i];
    int pnow = 0;
    for (int j = 2; B[i][j] != 0; ++j) {
      while (pnow != 0 && B[i][pnow + 1] != B[i][j]) pnow = P[i][pnow];
      if (B[i][pnow + 1] == B[i][j]) ++pnow;
      P[i][j] = pnow;
    }
  }
  for (int i = 1; A[i] != 0; ++i)
    for (int j = i; A[j] != 0; ++j) {
      SNOW.clear();
      for (int k = i; k <= j; ++k) {
        C[k - i + 1] = A[k];
        SNOW += A[k];
      }
      memset(prefix, 0, sizeof(prefix));
      int pnow = 0;
      for (int k = 2; k <= j - i + 1; ++k) {
        while (pnow != 0 && C[pnow + 1] != C[k]) pnow = prefix[pnow];
        if (C[pnow + 1] == C[k]) ++pnow;
        prefix[k] = pnow;
      }
      bool ok = true;
      for (int k = 1; k <= N; ++k) {
        int matchnow = 0, matches = 0;
        for (int l = 1; B[k][l] != 0; ++l) {
          while (matchnow != 0 && C[matchnow + 1] != B[k][l])
            matchnow = prefix[matchnow];
          if (C[matchnow + 1] == B[k][l]) ++matchnow;
          if (matchnow == j - i + 1) {
            ++matches;
            matchnow = prefix[matchnow];
          }
        }
        if (matches < L[k] || matches > R[k]) {
          ok = false;
          break;
        }
      }
      if (ok) M.insert(SNOW);
    }
  cout << M.size();
}

```