#include <bits/stdc++.h>
using namespace std;
int n, j, i, k, l[15], r[15];
int cur, e, q, t, x, y, ans, J, X, Y;
string s, p[15];
vector<pair<int, int> > v;
int main() {
  cin >> s;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> p[i] >> l[i] >> r[i];
  for (i = 0; i < s.size(); i++)
    for (j = i; j < s.size(); j++) {
      t = 1;
      for (q = 1; q <= n; q++) {
        cur = 0;
        for (k = 0; k <= (int)p[q].size() - j + i - 1; k++) {
          J = 1;
          for (e = 0; e < j - i + 1; e++)
            if (s[i + e] != p[q][k + e]) {
              J = 0;
              break;
            }
          if (J) cur++;
        }
        if (!(cur >= l[q] && cur <= r[q])) {
          t = 0;
          break;
        }
      }
      if (t) v.push_back(make_pair(i, j));
    }
  for (i = 0; i < v.size(); i++) {
    t = 1;
    x = v[i].first;
    y = v[i].second;
    for (j = 0; j < i; j++) {
      X = v[j].first;
      Y = v[j].second;
      if (Y - X != y - x) continue;
      e = 1;
      for (k = 0; k < Y - X + 1; k++)
        if (s[x + k] != s[X + k]) {
          e = 0;
          break;
        }
      if (e == 1) {
        t = 0;
        break;
      }
    }
    if (t) ans++;
  }
  cout << ans << endl;
}

### Prompt

Generate 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>
using namespace std;
int n, j, i, k, l[15], r[15];
int cur, e, q, t, x, y, ans, J, X, Y;
string s, p[15];
vector<pair<int, int> > v;
int main() {
  cin >> s;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> p[i] >> l[i] >> r[i];
  for (i = 0; i < s.size(); i++)
    for (j = i; j < s.size(); j++) {
      t = 1;
      for (q = 1; q <= n; q++) {
        cur = 0;
        for (k = 0; k <= (int)p[q].size() - j + i - 1; k++) {
          J = 1;
          for (e = 0; e < j - i + 1; e++)
            if (s[i + e] != p[q][k + e]) {
              J = 0;
              break;
            }
          if (J) cur++;
        }
        if (!(cur >= l[q] && cur <= r[q])) {
          t = 0;
          break;
        }
      }
      if (t) v.push_back(make_pair(i, j));
    }
  for (i = 0; i < v.size(); i++) {
    t = 1;
    x = v[i].first;
    y = v[i].second;
    for (j = 0; j < i; j++) {
      X = v[j].first;
      Y = v[j].second;
      if (Y - X != y - x) continue;
      e = 1;
      for (k = 0; k < Y - X + 1; k++)
        if (s[x + k] != s[X + k]) {
          e = 0;
          break;
        }
      if (e == 1) {
        t = 0;
        break;
      }
    }
    if (t) ans++;
  }
  cout << ans << endl;
}

```