#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const double eps = 1e-9;
char s[1000000];
const int md0 = 1e9 + 7;
const int md1 = 1e9 + 9;
int len0, n;
int len[11];
long long h0[1000000];
long long h1[1000000];
long long stp0[1000000];
long long stp1[1000000];
char ss[11][100000];
vector<long long> ls[11];
int l[11];
int r[11];
int cnt(vector<long long>& a, long long first) {
  return lower_bound((a).begin(), a.end(), first + 1) -
         lower_bound((a).begin(), a.end(), first);
}
int main() {
  scanf("%s", s);
  len0 = strlen(s);
  stp0[0] = stp1[0] = 1;
  stp0[1] = stp1[1] = 239;
  for (int i = 2; i <= 1000000; i++) {
    stp0[i] = (stp0[i - 1] * stp0[1]) % md0;
    stp1[i] = (stp1[i - 1] * stp1[1]) % md1;
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s %d %d", ss[i], &l[i], &r[i]);
    len[i] = strlen(ss[i]);
    h0[len[i]] = 0;
    h1[len[i]] = 0;
    for (int j = len[i] - 1; j >= 0; j--) {
      h0[j] = (h0[j + 1] * stp0[1] + ss[i][j]) % md0;
      h1[j] = (h1[j + 1] * stp1[1] + ss[i][j]) % md1;
    }
    for (int j = 0; j < len[i]; j++) {
      for (int k = j; k < len[i]; k++) {
        ls[i].push_back(
            (h0[j] - (h0[k + 1] * stp0[k - j + 1]) % md0 + md0) % md0 +
            (long long)md0 *
                ((h1[j] - (h1[k + 1] * stp1[k - j + 1]) % md1 + md1) % md1));
      }
    }
    sort((ls[i]).begin(), ls[i].end());
  }
  h0[len0] = 0;
  h1[len0] = 0;
  for (int i = len0 - 1; i >= 0; i--) {
    h0[i] = (h0[i + 1] * stp0[1] + s[i]) % md0;
    h1[i] = (h1[i + 1] * stp1[1] + s[i]) % md1;
  }
  set<long long> S;
  for (int j = 0; j < len0; j++) {
    for (int k = j; k < len0; k++) {
      long long ff =
          (h0[j] - (h0[k + 1] * stp0[k - j + 1]) % md0 + md0) % md0 +
          (long long)md0 *
              ((h1[j] - (h1[k + 1] * stp1[k - j + 1]) % md1 + md1) % md1);
      bool good = true;
      for (int i = 0; i < n; i++) {
        int gg = cnt(ls[i], ff);
        if (!(gg >= l[i] && gg <= r[i])) {
          good = false;
          break;
        }
      }
      if (good) {
        S.insert(ff);
      }
    }
  }
  cout << ((int)(S).size()) << endl;
  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>
using namespace std;
const int inf = (int)1e9;
const double eps = 1e-9;
char s[1000000];
const int md0 = 1e9 + 7;
const int md1 = 1e9 + 9;
int len0, n;
int len[11];
long long h0[1000000];
long long h1[1000000];
long long stp0[1000000];
long long stp1[1000000];
char ss[11][100000];
vector<long long> ls[11];
int l[11];
int r[11];
int cnt(vector<long long>& a, long long first) {
  return lower_bound((a).begin(), a.end(), first + 1) -
         lower_bound((a).begin(), a.end(), first);
}
int main() {
  scanf("%s", s);
  len0 = strlen(s);
  stp0[0] = stp1[0] = 1;
  stp0[1] = stp1[1] = 239;
  for (int i = 2; i <= 1000000; i++) {
    stp0[i] = (stp0[i - 1] * stp0[1]) % md0;
    stp1[i] = (stp1[i - 1] * stp1[1]) % md1;
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s %d %d", ss[i], &l[i], &r[i]);
    len[i] = strlen(ss[i]);
    h0[len[i]] = 0;
    h1[len[i]] = 0;
    for (int j = len[i] - 1; j >= 0; j--) {
      h0[j] = (h0[j + 1] * stp0[1] + ss[i][j]) % md0;
      h1[j] = (h1[j + 1] * stp1[1] + ss[i][j]) % md1;
    }
    for (int j = 0; j < len[i]; j++) {
      for (int k = j; k < len[i]; k++) {
        ls[i].push_back(
            (h0[j] - (h0[k + 1] * stp0[k - j + 1]) % md0 + md0) % md0 +
            (long long)md0 *
                ((h1[j] - (h1[k + 1] * stp1[k - j + 1]) % md1 + md1) % md1));
      }
    }
    sort((ls[i]).begin(), ls[i].end());
  }
  h0[len0] = 0;
  h1[len0] = 0;
  for (int i = len0 - 1; i >= 0; i--) {
    h0[i] = (h0[i + 1] * stp0[1] + s[i]) % md0;
    h1[i] = (h1[i + 1] * stp1[1] + s[i]) % md1;
  }
  set<long long> S;
  for (int j = 0; j < len0; j++) {
    for (int k = j; k < len0; k++) {
      long long ff =
          (h0[j] - (h0[k + 1] * stp0[k - j + 1]) % md0 + md0) % md0 +
          (long long)md0 *
              ((h1[j] - (h1[k + 1] * stp1[k - j + 1]) % md1 + md1) % md1);
      bool good = true;
      for (int i = 0; i < n; i++) {
        int gg = cnt(ls[i], ff);
        if (!(gg >= l[i] && gg <= r[i])) {
          good = false;
          break;
        }
      }
      if (good) {
        S.insert(ff);
      }
    }
  }
  cout << ((int)(S).size()) << endl;
  return 0;
}

```