#include <bits/stdc++.h>
using namespace std;
struct query {
  char s[2010];
  int fail[2010];
  int l, r;
} q[20];
char s2[2010];
int fail[2010];
void init(int k) {
  int i = 0, j;
  q[k].fail[0] = -1;
  int lp = strlen(q[k].s);
  for (j = 1; j < lp; j++) {
    for (i = q[k].fail[j - 1]; i >= 0 && q[k].s[i + 1] != q[k].s[j];
         i = q[k].fail[i])
      ;
    q[k].fail[j] = (q[k].s[i + 1] == q[k].s[j] ? i + 1 : -1);
  }
}
void init2() {
  int i = 0, j;
  fail[0] = -1;
  int lp = strlen(s2);
  for (j = 1; j < lp; j++) {
    for (i = fail[j - 1]; i >= 0 && s2[i + 1] != s2[j]; i = fail[i])
      ;
    fail[j] = (s2[i + 1] == s2[j] ? i + 1 : -1);
  }
}
bool check(int k) {
  int f[2010];
  int now = -1;
  int len = strlen(q[k].s);
  for (int i = 0; i < len; i++) {
    while (now != -1 && q[k].s[i] != s2[now + 1]) now = fail[now];
    f[i] = (q[k].s[i] == s2[now + 1] ? now + 1 : -1);
    now = f[i];
  }
  int total = 0;
  int len2 = strlen(s2);
  for (int i = 0; i < len; i++)
    if (f[i] == len2 - 1) total++;
  if (total >= q[k].l && total <= q[k].r)
    return true;
  else
    return false;
}
const int kind = 26;
struct node {
  node *fail;
  node *next[kind];
  int count;
  node() {
    fail = NULL;
    count = 0;
    memset(next, NULL, sizeof(next));
  }
};
int total = 0;
void insert(char *str, node *root) {
  node *p = root;
  int i = 0, index;
  while (str[i]) {
    index = str[i] - 'a';
    if (p->next[index] == NULL) p->next[index] = new node();
    p = p->next[index];
    i++;
  }
  if (p->count == 0) total++;
  p->count++;
}
int main() {
  char s[2010];
  scanf("%s", s);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s %d%d", q[i].s, &q[i].l, &q[i].r);
    int len = strlen(q[i].s);
    q[i].s[len] = '!';
    q[i].s[len + 1] = '\0';
    init(i);
  }
  int len = strlen(s);
  node *root = new node();
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) {
      int len = j - i + 1;
      for (int k = 0; k < len; k++) s2[k] = s[i + k];
      s2[len] = '\0';
      init2();
      bool flag = true;
      for (int k = 0; k < n; k++)
        if (!check(k)) flag = false;
      if (flag) {
        insert(s2, root);
      }
    }
  printf("%d\n", total);
  return 0;
}

### Prompt

Please formulate 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;
struct query {
  char s[2010];
  int fail[2010];
  int l, r;
} q[20];
char s2[2010];
int fail[2010];
void init(int k) {
  int i = 0, j;
  q[k].fail[0] = -1;
  int lp = strlen(q[k].s);
  for (j = 1; j < lp; j++) {
    for (i = q[k].fail[j - 1]; i >= 0 && q[k].s[i + 1] != q[k].s[j];
         i = q[k].fail[i])
      ;
    q[k].fail[j] = (q[k].s[i + 1] == q[k].s[j] ? i + 1 : -1);
  }
}
void init2() {
  int i = 0, j;
  fail[0] = -1;
  int lp = strlen(s2);
  for (j = 1; j < lp; j++) {
    for (i = fail[j - 1]; i >= 0 && s2[i + 1] != s2[j]; i = fail[i])
      ;
    fail[j] = (s2[i + 1] == s2[j] ? i + 1 : -1);
  }
}
bool check(int k) {
  int f[2010];
  int now = -1;
  int len = strlen(q[k].s);
  for (int i = 0; i < len; i++) {
    while (now != -1 && q[k].s[i] != s2[now + 1]) now = fail[now];
    f[i] = (q[k].s[i] == s2[now + 1] ? now + 1 : -1);
    now = f[i];
  }
  int total = 0;
  int len2 = strlen(s2);
  for (int i = 0; i < len; i++)
    if (f[i] == len2 - 1) total++;
  if (total >= q[k].l && total <= q[k].r)
    return true;
  else
    return false;
}
const int kind = 26;
struct node {
  node *fail;
  node *next[kind];
  int count;
  node() {
    fail = NULL;
    count = 0;
    memset(next, NULL, sizeof(next));
  }
};
int total = 0;
void insert(char *str, node *root) {
  node *p = root;
  int i = 0, index;
  while (str[i]) {
    index = str[i] - 'a';
    if (p->next[index] == NULL) p->next[index] = new node();
    p = p->next[index];
    i++;
  }
  if (p->count == 0) total++;
  p->count++;
}
int main() {
  char s[2010];
  scanf("%s", s);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s %d%d", q[i].s, &q[i].l, &q[i].r);
    int len = strlen(q[i].s);
    q[i].s[len] = '!';
    q[i].s[len + 1] = '\0';
    init(i);
  }
  int len = strlen(s);
  node *root = new node();
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) {
      int len = j - i + 1;
      for (int k = 0; k < len; k++) s2[k] = s[i + k];
      s2[len] = '\0';
      init2();
      bool flag = true;
      for (int k = 0; k < n; k++)
        if (!check(k)) flag = false;
      if (flag) {
        insert(s2, root);
      }
    }
  printf("%d\n", total);
  return 0;
}

```