#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100001;
struct KMP {
  int n;
  char pat[MAXN];
  vector<int> fail;
  void init(const char* s, int _n = -1) {
    n = _n == -1 ? strlen(s) : _n;
    memcpy(pat, s, sizeof(char) * (n + 1));
    fail = vector<int>(n + 1);
    for (int i = 0, j = fail[0] = -1; i < n; fail[++i] = ++j) {
      for (; j >= 0 && s[i] != s[j]; j = fail[j])
        ;
    }
  }
  int match(const char* s, int m = -1) {
    int ret = 0;
    m = m == -1 ? strlen(s) : m;
    for (int i = 0, j = 0; i < m; ++i) {
      for (; j > 0 && s[i] != pat[j]; j = fail[j])
        ;
      if (s[i] == pat[j]) {
        ++j;
      }
      if (j == n) {
        j = fail[j];
        ++ret;
      }
    }
    return ret;
  }
} kmp;
int n;
string s;
string p[10];
int l[10], r[10];
char buf[65536];
int main() {
  scanf("%s%d", buf, &n);
  s = string(buf);
  for (int i = 0; i < n; ++i) {
    scanf("%s%d%d", buf, &l[i], &r[i]);
    p[i] = string(buf);
  }
  int res = 0;
  for (int i = 1; i <= (int)s.length(); ++i) {
    set<string> st;
    for (int j = 0; j + i <= (int)s.length(); ++j) {
      string t = s.substr(j, i);
      if (st.find(t) != st.end()) {
        continue;
      }
      st.insert(t);
      kmp.init(t.c_str());
      bool ok = true;
      for (int k = 0; k < n && ok; ++k) {
        int count = kmp.match(p[k].c_str());
        ok = (count >= l[k] && count <= r[k]);
      }
      if (ok) {
        ++res;
      }
    }
  }
  printf("%d\n", res);
  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 = 100001;
struct KMP {
  int n;
  char pat[MAXN];
  vector<int> fail;
  void init(const char* s, int _n = -1) {
    n = _n == -1 ? strlen(s) : _n;
    memcpy(pat, s, sizeof(char) * (n + 1));
    fail = vector<int>(n + 1);
    for (int i = 0, j = fail[0] = -1; i < n; fail[++i] = ++j) {
      for (; j >= 0 && s[i] != s[j]; j = fail[j])
        ;
    }
  }
  int match(const char* s, int m = -1) {
    int ret = 0;
    m = m == -1 ? strlen(s) : m;
    for (int i = 0, j = 0; i < m; ++i) {
      for (; j > 0 && s[i] != pat[j]; j = fail[j])
        ;
      if (s[i] == pat[j]) {
        ++j;
      }
      if (j == n) {
        j = fail[j];
        ++ret;
      }
    }
    return ret;
  }
} kmp;
int n;
string s;
string p[10];
int l[10], r[10];
char buf[65536];
int main() {
  scanf("%s%d", buf, &n);
  s = string(buf);
  for (int i = 0; i < n; ++i) {
    scanf("%s%d%d", buf, &l[i], &r[i]);
    p[i] = string(buf);
  }
  int res = 0;
  for (int i = 1; i <= (int)s.length(); ++i) {
    set<string> st;
    for (int j = 0; j + i <= (int)s.length(); ++j) {
      string t = s.substr(j, i);
      if (st.find(t) != st.end()) {
        continue;
      }
      st.insert(t);
      kmp.init(t.c_str());
      bool ok = true;
      for (int k = 0; k < n && ok; ++k) {
        int count = kmp.match(p[k].c_str());
        ok = (count >= l[k] && count <= r[k]);
      }
      if (ok) {
        ++res;
      }
    }
  }
  printf("%d\n", res);
  return 0;
}

```