#include <bits/stdc++.h>
const double PI = acos(-1.0);
const double EPS = 1E-9;
const int INF = (int)1E9;
using namespace std;
int N;
char pp[11][1005];
int ll[11], rr[11];
char str[1005];
char pattern[1005];
char target[1005];
int nxt[1005];
void calcNext() {
  int k = 0;
  int len = strlen(pattern);
  nxt[0] = 0;
  for (int i = (1); i <= (len); i++) {
    if (pattern[i] == pattern[k])
      nxt[i] = nxt[k++];
    else {
      nxt[i] = k;
      while (k > 0 && pattern[i] != pattern[k]) k = nxt[k];
      if (pattern[k] == pattern[i]) k++;
    }
  }
}
int matchPattern() {
  int ans = 0;
  int i = 0, j = 0;
  int lenp = strlen(pattern);
  int lent = strlen(target);
  while (i < lenp && j < lent) {
    if (target[j] == pattern[i]) {
      i++;
      j++;
      if (i == lenp) {
        ans++;
        i = nxt[i];
      }
    } else {
      if (i)
        i = nxt[i];
      else
        j++;
    }
  }
  return ans;
}
int L;
set<string> anss;
int main() {
  scanf("%s", str);
  L = strlen(str);
  cin >> N;
  for (int i = 0; i < (N); i++) {
    scanf("%s%d%d", pp[i], &ll[i], &rr[i]);
  }
  anss.clear();
  for (int i = 0; i < (L); i++) {
    for (int l = (1); l <= (L - i); l++) {
      bool ok = 1;
      if (anss.find(string(str + i, str + i + l)) != anss.end()) continue;
      strcpy(pattern, str + i);
      pattern[l] = 0;
      calcNext();
      for (int k = 0; k < (N); k++) {
        strcpy(target, pp[k]);
        int cnt = matchPattern();
        if (cnt < ll[k] || cnt > rr[k]) {
          ok = 0;
          break;
        }
      }
      if (ok) {
        anss.insert(string(str + i, str + i + l));
      }
    }
  }
  cout << anss.size() << endl;
  set<string>::iterator r;
  return 0;
}

### Prompt

Your challenge is to write 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>
const double PI = acos(-1.0);
const double EPS = 1E-9;
const int INF = (int)1E9;
using namespace std;
int N;
char pp[11][1005];
int ll[11], rr[11];
char str[1005];
char pattern[1005];
char target[1005];
int nxt[1005];
void calcNext() {
  int k = 0;
  int len = strlen(pattern);
  nxt[0] = 0;
  for (int i = (1); i <= (len); i++) {
    if (pattern[i] == pattern[k])
      nxt[i] = nxt[k++];
    else {
      nxt[i] = k;
      while (k > 0 && pattern[i] != pattern[k]) k = nxt[k];
      if (pattern[k] == pattern[i]) k++;
    }
  }
}
int matchPattern() {
  int ans = 0;
  int i = 0, j = 0;
  int lenp = strlen(pattern);
  int lent = strlen(target);
  while (i < lenp && j < lent) {
    if (target[j] == pattern[i]) {
      i++;
      j++;
      if (i == lenp) {
        ans++;
        i = nxt[i];
      }
    } else {
      if (i)
        i = nxt[i];
      else
        j++;
    }
  }
  return ans;
}
int L;
set<string> anss;
int main() {
  scanf("%s", str);
  L = strlen(str);
  cin >> N;
  for (int i = 0; i < (N); i++) {
    scanf("%s%d%d", pp[i], &ll[i], &rr[i]);
  }
  anss.clear();
  for (int i = 0; i < (L); i++) {
    for (int l = (1); l <= (L - i); l++) {
      bool ok = 1;
      if (anss.find(string(str + i, str + i + l)) != anss.end()) continue;
      strcpy(pattern, str + i);
      pattern[l] = 0;
      calcNext();
      for (int k = 0; k < (N); k++) {
        strcpy(target, pp[k]);
        int cnt = matchPattern();
        if (cnt < ll[k] || cnt > rr[k]) {
          ok = 0;
          break;
        }
      }
      if (ok) {
        anss.insert(string(str + i, str + i + l));
      }
    }
  }
  cout << anss.size() << endl;
  set<string>::iterator r;
  return 0;
}

```