#include <bits/stdc++.h>
using namespace std;
char p[15][2100];
int l[15];
int r[15];
char s[2100];
int n;
vector<int> pfunc(string s, string t) {
  string res = s + '$' + t;
  vector<int> p(res.size());
  for (int i = 1; i < (int)res.size(); i++) {
    int j = p[i - 1];
    while (j && res[j] != res[i]) {
      j = p[j - 1];
    }
    if (res[j] == res[i]) j++;
    p[i] = j;
  }
  vector<int> cnt(s.size() + 1);
  for (int i = (int)s.size() + 1; i < (int)res.size(); i++) {
    assert(p[i] <= (int)s.size());
    cnt[p[i]]++;
  }
  for (int i = (int)s.size(); i > 0; i--) cnt[p[i - 1]] += cnt[i];
  return cnt;
}
vector<int> cnts[15];
bool check(int len) {
  for (int i = 0; i < n; i++)
    if (!(l[i] <= cnts[i][len] && cnts[i][len] <= r[i])) return 0;
  return 1;
}
int lcp[2100];
int main() {
  scanf("%s", s);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%s %d %d", p[i], &l[i], &r[i]);
  int ans = 0;
  int slen = strlen(s);
  for (int i = slen - 1; i >= 0; i--) {
    for (int j = 0; j < n; j++) cnts[j] = pfunc(s + i, p[j]);
    for (int j = 0; j < slen; j++) lcp[j] = (s[i] == s[j]) ? lcp[j + 1] + 1 : 0;
    for (int len = (i ? *max_element(lcp, lcp + i) : 0) + 1; s[i + len - 1];
         len++) {
      ans += check(len);
    }
  }
  printf("%d\n", ans);
  ;
  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;
char p[15][2100];
int l[15];
int r[15];
char s[2100];
int n;
vector<int> pfunc(string s, string t) {
  string res = s + '$' + t;
  vector<int> p(res.size());
  for (int i = 1; i < (int)res.size(); i++) {
    int j = p[i - 1];
    while (j && res[j] != res[i]) {
      j = p[j - 1];
    }
    if (res[j] == res[i]) j++;
    p[i] = j;
  }
  vector<int> cnt(s.size() + 1);
  for (int i = (int)s.size() + 1; i < (int)res.size(); i++) {
    assert(p[i] <= (int)s.size());
    cnt[p[i]]++;
  }
  for (int i = (int)s.size(); i > 0; i--) cnt[p[i - 1]] += cnt[i];
  return cnt;
}
vector<int> cnts[15];
bool check(int len) {
  for (int i = 0; i < n; i++)
    if (!(l[i] <= cnts[i][len] && cnts[i][len] <= r[i])) return 0;
  return 1;
}
int lcp[2100];
int main() {
  scanf("%s", s);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%s %d %d", p[i], &l[i], &r[i]);
  int ans = 0;
  int slen = strlen(s);
  for (int i = slen - 1; i >= 0; i--) {
    for (int j = 0; j < n; j++) cnts[j] = pfunc(s + i, p[j]);
    for (int j = 0; j < slen; j++) lcp[j] = (s[i] == s[j]) ? lcp[j + 1] + 1 : 0;
    for (int len = (i ? *max_element(lcp, lcp + i) : 0) + 1; s[i + len - 1];
         len++) {
      ans += check(len);
    }
  }
  printf("%d\n", ans);
  ;
  return 0;
}

```