#include <bits/stdc++.h>
using namespace std;
char str[200], pat[20][200];
map<pair<int, int>, int> H[20];
set<pair<int, int> > st;
int a[20], b[20];
int N;
int main() {
  int i, j, k, h, len, ans, cnt;
  scanf("%s", str);
  scanf("%d", &N);
  for (i = 0; i < N; i++) scanf("%s%d%d", &pat[i], &a[i], &b[i]);
  for (i = 0; i < N; i++) {
    len = strlen(pat[i]);
    for (j = 0; j < len; j++) {
      h = 0;
      for (k = j; k < len; k++) {
        h = h * 129 + pat[i][k];
        if (!H[i].count(pair<int, int>(h, k - j + 1)))
          H[i][pair<int, int>(h, k - j + 1)] = 0;
        H[i][pair<int, int>(h, k - j + 1)]++;
      }
    }
  }
  ans = 0;
  len = strlen(str);
  for (i = 0; i < len; i++) {
    h = 0;
    for (j = i; j < len; j++) {
      h = h * 129 + str[j];
      if (st.find(pair<int, int>(h, j - i + 1)) != st.end()) continue;
      st.insert(pair<int, int>(h, j - i + 1));
      for (k = 0; k < N; k++) {
        if (!H[k].count(pair<int, int>(h, j - i + 1)))
          cnt = 0;
        else
          cnt = H[k][pair<int, int>(h, j - i + 1)];
        if (cnt < a[k] || cnt > b[k]) break;
      }
      if (k == N) ans++;
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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;
char str[200], pat[20][200];
map<pair<int, int>, int> H[20];
set<pair<int, int> > st;
int a[20], b[20];
int N;
int main() {
  int i, j, k, h, len, ans, cnt;
  scanf("%s", str);
  scanf("%d", &N);
  for (i = 0; i < N; i++) scanf("%s%d%d", &pat[i], &a[i], &b[i]);
  for (i = 0; i < N; i++) {
    len = strlen(pat[i]);
    for (j = 0; j < len; j++) {
      h = 0;
      for (k = j; k < len; k++) {
        h = h * 129 + pat[i][k];
        if (!H[i].count(pair<int, int>(h, k - j + 1)))
          H[i][pair<int, int>(h, k - j + 1)] = 0;
        H[i][pair<int, int>(h, k - j + 1)]++;
      }
    }
  }
  ans = 0;
  len = strlen(str);
  for (i = 0; i < len; i++) {
    h = 0;
    for (j = i; j < len; j++) {
      h = h * 129 + str[j];
      if (st.find(pair<int, int>(h, j - i + 1)) != st.end()) continue;
      st.insert(pair<int, int>(h, j - i + 1));
      for (k = 0; k < N; k++) {
        if (!H[k].count(pair<int, int>(h, j - i + 1)))
          cnt = 0;
        else
          cnt = H[k][pair<int, int>(h, j - i + 1)];
        if (cnt < a[k] || cnt > b[k]) break;
      }
      if (k == N) ans++;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```