#include <bits/stdc++.h>
int n, m, ans;
char s[209];
char f[19][209];
int g[19][4];
int a[40009][2], b;
int main() {
  scanf("%s", s);
  scanf("%d", &n);
  for (int h = 0; h < n; h++) {
    scanf("%s %d %d", f[h], &g[h][0], &g[h][1]);
    g[h][2] = strlen(f[h]);
  }
  m = strlen(s);
  for (int h = 0; h < m; h++)
    for (int j = h; j < m; j++) {
      int v = 0;
      for (int i = 0; i < n; i++) {
        int k = 0, t = j - h + 1;
        for (int l = 0; l <= g[i][2] - t; l++) {
          bool p = 0;
          for (int r = l; r < l + t; r++)
            if (s[h + r - l] != f[i][r]) {
              p = 1;
              break;
            }
          if (!p) k++;
        }
        if (k >= g[i][0] && k <= g[i][1])
          v++;
        else
          break;
      }
      if (v == n) a[b][0] = h, a[b++][1] = j;
    }
  for (int h = 0; h < b; h++) {
    int p = 1;
    for (int j = 0; j < h; j++) {
      if (a[h][1] - a[h][0] != a[j][1] - a[j][0]) continue;
      bool w = 0;
      for (int i = 0; i < a[h][1] - a[h][0] + 1; i++)
        if (s[a[h][0] + i] != s[a[j][0] + i]) {
          w = 1;
          break;
        }
      if (!w) {
        p = 0;
        break;
      }
    }
    if (p) ans++;
  }
  printf("%d\n", ans);
}

### 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>
int n, m, ans;
char s[209];
char f[19][209];
int g[19][4];
int a[40009][2], b;
int main() {
  scanf("%s", s);
  scanf("%d", &n);
  for (int h = 0; h < n; h++) {
    scanf("%s %d %d", f[h], &g[h][0], &g[h][1]);
    g[h][2] = strlen(f[h]);
  }
  m = strlen(s);
  for (int h = 0; h < m; h++)
    for (int j = h; j < m; j++) {
      int v = 0;
      for (int i = 0; i < n; i++) {
        int k = 0, t = j - h + 1;
        for (int l = 0; l <= g[i][2] - t; l++) {
          bool p = 0;
          for (int r = l; r < l + t; r++)
            if (s[h + r - l] != f[i][r]) {
              p = 1;
              break;
            }
          if (!p) k++;
        }
        if (k >= g[i][0] && k <= g[i][1])
          v++;
        else
          break;
      }
      if (v == n) a[b][0] = h, a[b++][1] = j;
    }
  for (int h = 0; h < b; h++) {
    int p = 1;
    for (int j = 0; j < h; j++) {
      if (a[h][1] - a[h][0] != a[j][1] - a[j][0]) continue;
      bool w = 0;
      for (int i = 0; i < a[h][1] - a[h][0] + 1; i++)
        if (s[a[h][0] + i] != s[a[j][0] + i]) {
          w = 1;
          break;
        }
      if (!w) {
        p = 0;
        break;
      }
    }
    if (p) ans++;
  }
  printf("%d\n", ans);
}

```