#include <bits/stdc++.h>
using namespace std;
vector<long long> good;
long long n, r;
struct R {
  string str;
  long long l, r;
};
R rule[15];
long long pow(long long x, long long p, long long m) {
  long long res = 1;
  if (p == 0) return 1;
  long long half = pow(x, p / 2, m);
  half *= half;
  half %= m;
  if (p & 1) half *= x;
  half += m;
  half %= m;
  return half;
}
bool ok(long long x, long long len, long long rulenum) {
  long long cnt = 0;
  string &str = rule[rulenum].str;
  if (str.size() < len) {
    return rule[rulenum].l <= cnt && cnt <= rule[rulenum].r;
  };
  long long p = 0;
  for (long long i = 0; i < len; i++) {
    p *= 100000007;
    p %= 1000000009;
    p += str[i] - 'a' + 1;
    p %= 1000000009;
  }
  long long minu = pow(100000007, len, 1000000009);
  for (long long i = 0; i + len <= str.size(); i++) {
    if (p == x) cnt++;
    if (i + len == str.size()) break;
    p *= 100000007;
    p %= 1000000009;
    p += str[i + len] - 'a' + 1;
    p -= (str[i] - 'a' + 1) * minu;
    p %= 1000000009;
    p += 1000000009;
    p %= 1000000009;
  }
  return rule[rulenum].l <= cnt && cnt <= rule[rulenum].r;
}
bool able(long long x, long long len) {
  for (long long i = 0; i < r; i++) {
    if (!ok(x, len, i)) return false;
  }
  return true;
}
int main() {
  string str;
  cin >> str;
  n = str.size();
  cin >> r;
  for (long long i = 0; i < r; i++) {
    cin >> rule[i].str >> rule[i].l >> rule[i].r;
  }
  for (long long i = 1; i <= n; i++) {
    long long x = 0;
    for (long long j = 0; j < i; j++) {
      x *= 100000007;
      x %= 1000000009;
      x += str[j] - 'a' + 1;
      x %= 1000000009;
    }
    long long minu = pow(100000007, i, 1000000009);
    for (long long j = 0; j + i <= n; j++) {
      if (able(x, i)) good.push_back(x);
      if (j + i == n) break;
      x *= 100000007;
      x %= 1000000009;
      x += str[j + i] - 'a' + 1;
      x -= (str[j] - 'a' + 1) * minu;
      x %= 1000000009;
      x += 1000000009;
      x %= 1000000009;
    }
  }
  sort(good.begin(), good.end());
  good.erase(unique(good.begin(), good.end()), good.end());
  cout << good.size() << endl;
  return 0;
}

### 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;
vector<long long> good;
long long n, r;
struct R {
  string str;
  long long l, r;
};
R rule[15];
long long pow(long long x, long long p, long long m) {
  long long res = 1;
  if (p == 0) return 1;
  long long half = pow(x, p / 2, m);
  half *= half;
  half %= m;
  if (p & 1) half *= x;
  half += m;
  half %= m;
  return half;
}
bool ok(long long x, long long len, long long rulenum) {
  long long cnt = 0;
  string &str = rule[rulenum].str;
  if (str.size() < len) {
    return rule[rulenum].l <= cnt && cnt <= rule[rulenum].r;
  };
  long long p = 0;
  for (long long i = 0; i < len; i++) {
    p *= 100000007;
    p %= 1000000009;
    p += str[i] - 'a' + 1;
    p %= 1000000009;
  }
  long long minu = pow(100000007, len, 1000000009);
  for (long long i = 0; i + len <= str.size(); i++) {
    if (p == x) cnt++;
    if (i + len == str.size()) break;
    p *= 100000007;
    p %= 1000000009;
    p += str[i + len] - 'a' + 1;
    p -= (str[i] - 'a' + 1) * minu;
    p %= 1000000009;
    p += 1000000009;
    p %= 1000000009;
  }
  return rule[rulenum].l <= cnt && cnt <= rule[rulenum].r;
}
bool able(long long x, long long len) {
  for (long long i = 0; i < r; i++) {
    if (!ok(x, len, i)) return false;
  }
  return true;
}
int main() {
  string str;
  cin >> str;
  n = str.size();
  cin >> r;
  for (long long i = 0; i < r; i++) {
    cin >> rule[i].str >> rule[i].l >> rule[i].r;
  }
  for (long long i = 1; i <= n; i++) {
    long long x = 0;
    for (long long j = 0; j < i; j++) {
      x *= 100000007;
      x %= 1000000009;
      x += str[j] - 'a' + 1;
      x %= 1000000009;
    }
    long long minu = pow(100000007, i, 1000000009);
    for (long long j = 0; j + i <= n; j++) {
      if (able(x, i)) good.push_back(x);
      if (j + i == n) break;
      x *= 100000007;
      x %= 1000000009;
      x += str[j + i] - 'a' + 1;
      x -= (str[j] - 'a' + 1) * minu;
      x %= 1000000009;
      x += 1000000009;
      x %= 1000000009;
    }
  }
  sort(good.begin(), good.end());
  good.erase(unique(good.begin(), good.end()), good.end());
  cout << good.size() << endl;
  return 0;
}

```