#include <bits/stdc++.h>
using namespace std;
template <typename T>
int size(T &a) {
  return (int)a.size();
}
template <typename T>
T sqr(T a) {
  return a * a;
}
const int MAXN = 2 * int(1e3) + 2;
char s[MAXN];
char buf[MAXN];
struct node {
  int to[26];
  node() { memset((to), (-1), sizeof(to)); }
};
const int MAXTR = MAXN * MAXN / 2;
struct trie {
  node tr[MAXTR];
  char ok[MAXTR];
  int occur[MAXTR];
  int lastID;
  int l, r;
  trie() {
    lastID = 1;
    tr[0] = node();
    ok[0] = false;
  }
  void init() {
    for (int i = (0); i < (lastID); ++i) occur[i] = 0;
  }
  void clear() {
    lastID = 1;
    memset((tr[0].to), (-1), sizeof(tr[0].to));
    ok[0] = false;
  }
  void addWord(char *s) {
    int root = 0;
    while (s[0] != 0) {
      if (tr[root].to[s[0] - 'a'] == -1) {
        tr[root].to[s[0] - 'a'] = lastID;
        memset((tr[lastID].to), (-1), sizeof(tr[lastID].to));
        ok[lastID] = true;
        lastID++;
      }
      root = tr[root].to[s[0] - 'a'];
      s++;
    }
  }
  void processWord(char *s) {
    int root = 0;
    while (s[0] != 0) {
      if (tr[root].to[s[0] - 'a'] == -1) {
        return;
      }
      root = tr[root].to[s[0] - 'a'];
      s++;
      occur[root]++;
    }
  }
  bool isOk(int state) { return occur[state] >= l && occur[state] <= r; }
};
void makeTrie(trie &T, char *s) {
  while (s[0] != 0) {
    T.addWord(s);
    s++;
  }
}
trie ST;
int ans = 0;
int N;
void solve() {
  gets(s);
  makeTrie(ST, s);
  scanf("%d\n", &N);
  for (int i = (0); i < (N); ++i) {
    ST.init();
    scanf("%s %d %d\n", buf, &ST.l, &ST.r);
    char *s = buf;
    while (s[0] != 0) {
      ST.processWord(s);
      s++;
    }
    for (int i = (1); i < (ST.lastID); ++i) {
      ST.ok[i] = ST.ok[i] && ST.isOk(i);
    }
  }
  ans = 0;
  for (int i = (0); i < (ST.lastID); ++i) {
    if (ST.ok[i]) ans++;
  }
  printf("%d\n", ans);
}
int main() {
  solve();
  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;
template <typename T>
int size(T &a) {
  return (int)a.size();
}
template <typename T>
T sqr(T a) {
  return a * a;
}
const int MAXN = 2 * int(1e3) + 2;
char s[MAXN];
char buf[MAXN];
struct node {
  int to[26];
  node() { memset((to), (-1), sizeof(to)); }
};
const int MAXTR = MAXN * MAXN / 2;
struct trie {
  node tr[MAXTR];
  char ok[MAXTR];
  int occur[MAXTR];
  int lastID;
  int l, r;
  trie() {
    lastID = 1;
    tr[0] = node();
    ok[0] = false;
  }
  void init() {
    for (int i = (0); i < (lastID); ++i) occur[i] = 0;
  }
  void clear() {
    lastID = 1;
    memset((tr[0].to), (-1), sizeof(tr[0].to));
    ok[0] = false;
  }
  void addWord(char *s) {
    int root = 0;
    while (s[0] != 0) {
      if (tr[root].to[s[0] - 'a'] == -1) {
        tr[root].to[s[0] - 'a'] = lastID;
        memset((tr[lastID].to), (-1), sizeof(tr[lastID].to));
        ok[lastID] = true;
        lastID++;
      }
      root = tr[root].to[s[0] - 'a'];
      s++;
    }
  }
  void processWord(char *s) {
    int root = 0;
    while (s[0] != 0) {
      if (tr[root].to[s[0] - 'a'] == -1) {
        return;
      }
      root = tr[root].to[s[0] - 'a'];
      s++;
      occur[root]++;
    }
  }
  bool isOk(int state) { return occur[state] >= l && occur[state] <= r; }
};
void makeTrie(trie &T, char *s) {
  while (s[0] != 0) {
    T.addWord(s);
    s++;
  }
}
trie ST;
int ans = 0;
int N;
void solve() {
  gets(s);
  makeTrie(ST, s);
  scanf("%d\n", &N);
  for (int i = (0); i < (N); ++i) {
    ST.init();
    scanf("%s %d %d\n", buf, &ST.l, &ST.r);
    char *s = buf;
    while (s[0] != 0) {
      ST.processWord(s);
      s++;
    }
    for (int i = (1); i < (ST.lastID); ++i) {
      ST.ok[i] = ST.ok[i] && ST.isOk(i);
    }
  }
  ans = 0;
  for (int i = (0); i < (ST.lastID); ++i) {
    if (ST.ok[i]) ans++;
  }
  printf("%d\n", ans);
}
int main() {
  solve();
  return 0;
}

```