#include <bits/stdc++.h>
using namespace std;
const int K1 = 1000003;
const int K2 = 1000007;
const int MAXN = 50000;
long long B1[MAXN + 1], B2[MAXN + 1];
int n;
string s;
string p[10];
int l[10], r[10];
char buf[65536];
map<long long, int> gao(const string& s, int len) {
  long long v1 = 0;
  for (int j = 0; j < len && j < (int)s.length(); ++j) {
    v1 = v1 * K1 + s[j];
  }
  map<long long, int> mp;
  mp[v1] += 1;
  for (int j = len; j < (int)s.length(); ++j) {
    v1 = (v1 - B1[len - 1] * s[j - len]) * K1 + s[j];
    mp[v1] += 1;
  }
  return mp;
}
int gao(int len) {
  map<long long, int> ms = gao(s, len);
  vector<long long> vs;
  for (map<long long, int>::const_iterator i = ms.begin(); i != ms.end(); ++i) {
    vs.push_back(i->first);
  }
  for (int i = 0; i < n; ++i) {
    map<long long, int> mp = gao(p[i], len);
    vector<long long> newVS;
    for (int j = 0; j < (int)vs.size(); ++j) {
      int count = mp.find(vs[j]) == mp.end() ? 0 : mp[vs[j]];
      if (count >= l[i] && count <= r[i]) {
        newVS.push_back(vs[j]);
      }
    }
    vs.swap(newVS);
  }
  return (int)vs.size();
}
int main() {
  B1[0] = B2[0] = 1;
  for (int i = 1; i <= MAXN; ++i) {
    B1[i] = B1[i - 1] * K1;
    B2[i] = B2[i - 1] * K2;
  }
  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) {
    int cur = gao(i);
    res += cur;
  }
  printf("%d\n", res);
  return 0;
}

### Prompt

Your task is to create 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;
const int K1 = 1000003;
const int K2 = 1000007;
const int MAXN = 50000;
long long B1[MAXN + 1], B2[MAXN + 1];
int n;
string s;
string p[10];
int l[10], r[10];
char buf[65536];
map<long long, int> gao(const string& s, int len) {
  long long v1 = 0;
  for (int j = 0; j < len && j < (int)s.length(); ++j) {
    v1 = v1 * K1 + s[j];
  }
  map<long long, int> mp;
  mp[v1] += 1;
  for (int j = len; j < (int)s.length(); ++j) {
    v1 = (v1 - B1[len - 1] * s[j - len]) * K1 + s[j];
    mp[v1] += 1;
  }
  return mp;
}
int gao(int len) {
  map<long long, int> ms = gao(s, len);
  vector<long long> vs;
  for (map<long long, int>::const_iterator i = ms.begin(); i != ms.end(); ++i) {
    vs.push_back(i->first);
  }
  for (int i = 0; i < n; ++i) {
    map<long long, int> mp = gao(p[i], len);
    vector<long long> newVS;
    for (int j = 0; j < (int)vs.size(); ++j) {
      int count = mp.find(vs[j]) == mp.end() ? 0 : mp[vs[j]];
      if (count >= l[i] && count <= r[i]) {
        newVS.push_back(vs[j]);
      }
    }
    vs.swap(newVS);
  }
  return (int)vs.size();
}
int main() {
  B1[0] = B2[0] = 1;
  for (int i = 1; i <= MAXN; ++i) {
    B1[i] = B1[i - 1] * K1;
    B2[i] = B2[i - 1] * K2;
  }
  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) {
    int cur = gao(i);
    res += cur;
  }
  printf("%d\n", res);
  return 0;
}

```