#include <bits/stdc++.h>
using namespace std;
string s, str;
const int key = 239;
const int base = 1000000007;
set<int> ansset[2222];
long long hs[2222];
map<int, int> mp[11][2222];
int l[11], r[11];
int k;
bool pr(int h, int len) {
  for (int i = (1); i <= (k); i++) {
    int t = 0;
    t = mp[i][len][h];
    if ((t < l[i]) || (t > r[i])) {
      return false;
    }
  }
  return true;
}
int main() {
  cin >> s;
  int n = s.length();
  cin >> k;
  for (int tt = (1); tt <= (k); tt++) {
    cin >> str >> l[tt] >> r[tt];
    int len = str.size();
    for (int i = (0); i <= (len - 1); i++) {
      long long h = 0;
      for (int j = (i); j <= (len - 1); j++) {
        h = (h * key + str[j]) % base;
        mp[tt][j - i + 1][h]++;
      }
    }
  }
  for (int i = (0); i <= (n - 1); i++) {
    long long h = 0;
    for (int j = (i); j <= (n - 1); j++) {
      h = (h * key + s[j]) % base;
      if (pr(h, j - i + 1)) {
        ansset[j - i + 1].insert(h);
      }
    }
  }
  k = 0;
  for (int i = (1); i <= (n); i++) {
    k += ansset[i].size();
  }
  cout << k;
  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;
string s, str;
const int key = 239;
const int base = 1000000007;
set<int> ansset[2222];
long long hs[2222];
map<int, int> mp[11][2222];
int l[11], r[11];
int k;
bool pr(int h, int len) {
  for (int i = (1); i <= (k); i++) {
    int t = 0;
    t = mp[i][len][h];
    if ((t < l[i]) || (t > r[i])) {
      return false;
    }
  }
  return true;
}
int main() {
  cin >> s;
  int n = s.length();
  cin >> k;
  for (int tt = (1); tt <= (k); tt++) {
    cin >> str >> l[tt] >> r[tt];
    int len = str.size();
    for (int i = (0); i <= (len - 1); i++) {
      long long h = 0;
      for (int j = (i); j <= (len - 1); j++) {
        h = (h * key + str[j]) % base;
        mp[tt][j - i + 1][h]++;
      }
    }
  }
  for (int i = (0); i <= (n - 1); i++) {
    long long h = 0;
    for (int j = (i); j <= (n - 1); j++) {
      h = (h * key + s[j]) % base;
      if (pr(h, j - i + 1)) {
        ansset[j - i + 1].insert(h);
      }
    }
  }
  k = 0;
  for (int i = (1); i <= (n); i++) {
    k += ansset[i].size();
  }
  cout << k;
  return 0;
}

```