#include <bits/stdc++.h>
using namespace std;
int len[16], l[16], r[16], next[2048], slen, n, ans;
char s[2048], p[10][2048];
set<string> S;
string makeString(int a, int b) {
  string ret = "";
  for (int i = a; i <= b; i++) ret += s[i];
  return ret;
}
int main() {
  scanf("%s", s);
  slen = strlen(s);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s%d%d", p[i], &l[i], &r[i]);
    len[i] = strlen(p[i]);
  }
  for (int i = 0; i < slen; i++)
    for (int j = i; j < slen; j++) S.insert(makeString(i, j));
  for (set<string>::iterator it = S.begin(); it != S.end(); it++) {
    string curString = *it;
    int clen = curString.length();
    int f = 0;
    for (int i = 0; i < n; i++) {
      int occ = 0;
      if (clen <= len[i]) {
        for (int j = 0; j < len[i]; j++) {
          int ok = 1;
          for (int k = 0; k < clen; k++) {
            if (p[i][j + k] != curString[k]) {
              ok = 0;
              break;
            }
          }
          if (ok) occ++;
        }
      }
      if (occ >= l[i] && occ <= r[i]) f++;
    }
    if (f == n) {
      ans++;
    }
  }
  printf("%d\n", ans);
  return 0;
}

### 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;
int len[16], l[16], r[16], next[2048], slen, n, ans;
char s[2048], p[10][2048];
set<string> S;
string makeString(int a, int b) {
  string ret = "";
  for (int i = a; i <= b; i++) ret += s[i];
  return ret;
}
int main() {
  scanf("%s", s);
  slen = strlen(s);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s%d%d", p[i], &l[i], &r[i]);
    len[i] = strlen(p[i]);
  }
  for (int i = 0; i < slen; i++)
    for (int j = i; j < slen; j++) S.insert(makeString(i, j));
  for (set<string>::iterator it = S.begin(); it != S.end(); it++) {
    string curString = *it;
    int clen = curString.length();
    int f = 0;
    for (int i = 0; i < n; i++) {
      int occ = 0;
      if (clen <= len[i]) {
        for (int j = 0; j < len[i]; j++) {
          int ok = 1;
          for (int k = 0; k < clen; k++) {
            if (p[i][j + k] != curString[k]) {
              ok = 0;
              break;
            }
          }
          if (ok) occ++;
        }
      }
      if (occ >= l[i] && occ <= r[i]) f++;
    }
    if (f == n) {
      ans++;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```