#include <bits/stdc++.h>
using namespace std;
const int MaxN = 2020;
char s[MaxN];
int L;
int num, l[20], r[20];
char p[20][MaxN];
int pi[MaxN];
char T[MaxN], P[MaxN];
struct myString {
  int l, r;
  myString() {}
  myString(int _l, int _r) {
    l = _l;
    r = _r;
  }
};
bool cmp(const myString& A, const myString& B) {
  int la = A.r - A.l + 1;
  int lb = B.r - B.l + 1;
  if (la != lb) return (la < lb);
  for (int i = 0; i < la; i++) {
    if (s[A.l + i] != s[B.l + i]) return (s[A.l + i] < s[B.l + i]);
  }
  return false;
}
myString sol[MaxN * MaxN];
void ComputePrefixFunction(char* P, int m) {
  int q, k;
  pi[1] = 0;
  k = 0;
  for (q = 2; q <= m; q++) {
    while ((k > 0) && (P[k + 1] != P[q])) k = pi[k];
    if (P[k + 1] == P[q]) k++;
    pi[q] = k;
  }
}
bool ok(int left, int right, int num) {
  int n, m, i, q, total = 0;
  m = right - left + 1;
  for (int i = 1; i <= m; i++) P[i] = s[left + i - 1];
  n = strlen(p[num]);
  for (int i = 1; i <= n; i++) T[i] = p[num][i - 1];
  ComputePrefixFunction(P, m);
  q = 0;
  for (i = 1; i <= n; i++) {
    while ((q > 0) && (P[q + 1] != T[i])) q = pi[q];
    if (P[q + 1] == T[i]) q++;
    if (q == m) {
      total++;
      q = pi[q];
    }
  }
  return (total >= l[num] && total <= r[num]);
}
int main() {
  scanf("%s", s);
  L = strlen(s);
  scanf("%d", &num);
  for (int i = 1; i <= num; i++) {
    scanf("%s %d %d", p[i], &l[i], &r[i]);
  }
  int solNum = 0;
  for (int left = 0; left < L; left++)
    for (int right = left; right < L; right++) {
      bool Ok = true;
      for (int i = 1; i <= num && Ok; i++) {
        Ok = ok(left, right, i);
      }
      if (Ok) sol[solNum++] = myString(left, right);
    }
  if (solNum == 0) {
    printf("0\n");
    return 0;
  }
  int ret = solNum;
  sort(sol, sol + solNum, cmp);
  for (int i = 1; i < solNum; i++)
    if (!cmp(sol[i], sol[i - 1]) && !cmp(sol[i - 1], sol[i])) ret--;
  printf("%d\n", ret);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 MaxN = 2020;
char s[MaxN];
int L;
int num, l[20], r[20];
char p[20][MaxN];
int pi[MaxN];
char T[MaxN], P[MaxN];
struct myString {
  int l, r;
  myString() {}
  myString(int _l, int _r) {
    l = _l;
    r = _r;
  }
};
bool cmp(const myString& A, const myString& B) {
  int la = A.r - A.l + 1;
  int lb = B.r - B.l + 1;
  if (la != lb) return (la < lb);
  for (int i = 0; i < la; i++) {
    if (s[A.l + i] != s[B.l + i]) return (s[A.l + i] < s[B.l + i]);
  }
  return false;
}
myString sol[MaxN * MaxN];
void ComputePrefixFunction(char* P, int m) {
  int q, k;
  pi[1] = 0;
  k = 0;
  for (q = 2; q <= m; q++) {
    while ((k > 0) && (P[k + 1] != P[q])) k = pi[k];
    if (P[k + 1] == P[q]) k++;
    pi[q] = k;
  }
}
bool ok(int left, int right, int num) {
  int n, m, i, q, total = 0;
  m = right - left + 1;
  for (int i = 1; i <= m; i++) P[i] = s[left + i - 1];
  n = strlen(p[num]);
  for (int i = 1; i <= n; i++) T[i] = p[num][i - 1];
  ComputePrefixFunction(P, m);
  q = 0;
  for (i = 1; i <= n; i++) {
    while ((q > 0) && (P[q + 1] != T[i])) q = pi[q];
    if (P[q + 1] == T[i]) q++;
    if (q == m) {
      total++;
      q = pi[q];
    }
  }
  return (total >= l[num] && total <= r[num]);
}
int main() {
  scanf("%s", s);
  L = strlen(s);
  scanf("%d", &num);
  for (int i = 1; i <= num; i++) {
    scanf("%s %d %d", p[i], &l[i], &r[i]);
  }
  int solNum = 0;
  for (int left = 0; left < L; left++)
    for (int right = left; right < L; right++) {
      bool Ok = true;
      for (int i = 1; i <= num && Ok; i++) {
        Ok = ok(left, right, i);
      }
      if (Ok) sol[solNum++] = myString(left, right);
    }
  if (solNum == 0) {
    printf("0\n");
    return 0;
  }
  int ret = solNum;
  sort(sol, sol + solNum, cmp);
  for (int i = 1; i < solNum; i++)
    if (!cmp(sol[i], sol[i - 1]) && !cmp(sol[i - 1], sol[i])) ret--;
  printf("%d\n", ret);
  return 0;
}

```