#include <bits/stdc++.h>
using namespace std;
void preKmp(char *src, int m, int *rule) {
  int i = 0, j = -1;
  rule[0] = j;
  while (i < m) {
    while (j != -1 && src[j] != src[i]) j = rule[j];
    i++, j++;
    if (j >= m)
      rule[i] = rule[j - 1];
    else
      rule[i] = j;
  }
}
int recount = 0;
void kmp(char *src, int m, char *dest, int n) {
  int rule[10005];
  preKmp(src, m, rule);
  int i = 0, j = 0;
  while (i < n) {
    while (j != -1 && src[j] != dest[i]) {
      j = rule[j];
    }
    i++, j++;
    if (j == m) {
      recount++;
      j = rule[j];
    }
  }
}
char Reg[10005], ori[1000005];
char os[100][300], a[500], temp[500];
int l[100], r[100], n, len, le[100];
int main() {
  set<string> now;
  set<string>::iterator iter;
  bool flag;
  int ans = 0, i, j, k;
  while (scanf("%s", a) != EOF) {
    now.clear();
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
      scanf("%s%d%d", os[i], &l[i], &r[i]);
      le[i] = strlen(os[i]);
    }
    len = strlen(a);
    ans = 0;
    for (i = 0; i < len; i++)
      for (j = i; j < len; j++) {
        flag = true;
        for (k = 1; k <= n; k++) {
          recount = 0;
          kmp(a + i, j - i + 1, os[k], le[k]);
          if (recount < l[k] || recount > r[k]) {
            flag = false;
            break;
          }
        }
        if (flag) {
          for (k = i; k <= j; k++) temp[k - i] = a[k];
          temp[k - i] = '\0';
          string wc = temp;
          iter = now.find(wc);
          if (iter == now.end()) {
            now.insert(wc);
            ans++;
          }
        }
      }
    printf("%d\n", ans);
  }
}

### Prompt

Create a solution in CPP for 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;
void preKmp(char *src, int m, int *rule) {
  int i = 0, j = -1;
  rule[0] = j;
  while (i < m) {
    while (j != -1 && src[j] != src[i]) j = rule[j];
    i++, j++;
    if (j >= m)
      rule[i] = rule[j - 1];
    else
      rule[i] = j;
  }
}
int recount = 0;
void kmp(char *src, int m, char *dest, int n) {
  int rule[10005];
  preKmp(src, m, rule);
  int i = 0, j = 0;
  while (i < n) {
    while (j != -1 && src[j] != dest[i]) {
      j = rule[j];
    }
    i++, j++;
    if (j == m) {
      recount++;
      j = rule[j];
    }
  }
}
char Reg[10005], ori[1000005];
char os[100][300], a[500], temp[500];
int l[100], r[100], n, len, le[100];
int main() {
  set<string> now;
  set<string>::iterator iter;
  bool flag;
  int ans = 0, i, j, k;
  while (scanf("%s", a) != EOF) {
    now.clear();
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
      scanf("%s%d%d", os[i], &l[i], &r[i]);
      le[i] = strlen(os[i]);
    }
    len = strlen(a);
    ans = 0;
    for (i = 0; i < len; i++)
      for (j = i; j < len; j++) {
        flag = true;
        for (k = 1; k <= n; k++) {
          recount = 0;
          kmp(a + i, j - i + 1, os[k], le[k]);
          if (recount < l[k] || recount > r[k]) {
            flag = false;
            break;
          }
        }
        if (flag) {
          for (k = i; k <= j; k++) temp[k - i] = a[k];
          temp[k - i] = '\0';
          string wc = temp;
          iter = now.find(wc);
          if (iter == now.end()) {
            now.insert(wc);
            ans++;
          }
        }
      }
    printf("%d\n", ans);
  }
}

```