#include <bits/stdc++.h>
using namespace std;
const int N = 2000 + 5;
char an[N], str[N], rs[N][N];
int l[N], r[N];
set<string> aaa;
int main() {
  scanf("%s", str);
  int n, f, cnt, ans = 0, good = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", rs[i]);
    scanf("%d%d", &l[i], &r[i]);
  }
  f = strlen(str);
  for (int i = 0; i < f; ++i) {
    for (int j = i; j < f; ++j) {
      ans = 0;
      for (int k = 0; k < n; ++k) {
        cnt = 0;
        if ((j - i + 1) > strlen(rs[k]))
          ;
        else
          for (int m = 0; m < strlen(rs[k]) - (j - i); ++m) {
            if (strncmp(str + i, rs[k] + m, j - i + 1) == 0) cnt++;
          }
        if (l[k] <= cnt && cnt <= r[k]) ans++;
      }
      if (ans == n) {
        strncpy(an, str + i, j - i + 1);
        an[j - i + 1] = 0;
        aaa.insert(an);
      }
    }
  }
  cout << aaa.size() << endl;
}

### 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 int N = 2000 + 5;
char an[N], str[N], rs[N][N];
int l[N], r[N];
set<string> aaa;
int main() {
  scanf("%s", str);
  int n, f, cnt, ans = 0, good = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", rs[i]);
    scanf("%d%d", &l[i], &r[i]);
  }
  f = strlen(str);
  for (int i = 0; i < f; ++i) {
    for (int j = i; j < f; ++j) {
      ans = 0;
      for (int k = 0; k < n; ++k) {
        cnt = 0;
        if ((j - i + 1) > strlen(rs[k]))
          ;
        else
          for (int m = 0; m < strlen(rs[k]) - (j - i); ++m) {
            if (strncmp(str + i, rs[k] + m, j - i + 1) == 0) cnt++;
          }
        if (l[k] <= cnt && cnt <= r[k]) ans++;
      }
      if (ans == n) {
        strncpy(an, str + i, j - i + 1);
        an[j - i + 1] = 0;
        aaa.insert(an);
      }
    }
  }
  cout << aaa.size() << endl;
}

```