#include <bits/stdc++.h>
using namespace std;
string s[1000];
int a[1000], b[1000];
map<string, int> mp, Mp[1000];
set<string> Se;
int main() {
  int n, m;
  string S;
  cin >> S;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i] >> a[i] >> b[i];
    for (int I = 1; I <= s[i].length(); I++)
      for (int j = 0; I + j <= s[i].length(); j++) {
        string ss;
        for (int y = j; y < j + I; y++) ss += s[i][y];
        if (Mp[i].find(ss) == Mp[i].end())
          Mp[i][ss] = 1;
        else
          Mp[i][ss]++;
      }
  }
  long long res = 0;
  for (int i = 1; i <= S.length(); i++)
    for (int j = 0; i + j <= S.length(); j++) {
      string ss;
      for (int y = j; y < j + i; y++) ss += S[y];
      int flag = 1;
      for (int y = 0; y < n; y++) {
        if (Mp[y].find(ss) != Mp[y].end()) {
          if (Mp[y][ss] > b[y] || Mp[y][ss] < a[y]) {
            flag = 0;
          }
        } else if (a[y] > 0) {
          flag = 0;
        }
      }
      if (flag) {
        Se.insert(ss);
      }
    }
  cout << Se.size() << endl;
  cin >> n;
}

### Prompt

Create a solution in cpp for 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[1000];
int a[1000], b[1000];
map<string, int> mp, Mp[1000];
set<string> Se;
int main() {
  int n, m;
  string S;
  cin >> S;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i] >> a[i] >> b[i];
    for (int I = 1; I <= s[i].length(); I++)
      for (int j = 0; I + j <= s[i].length(); j++) {
        string ss;
        for (int y = j; y < j + I; y++) ss += s[i][y];
        if (Mp[i].find(ss) == Mp[i].end())
          Mp[i][ss] = 1;
        else
          Mp[i][ss]++;
      }
  }
  long long res = 0;
  for (int i = 1; i <= S.length(); i++)
    for (int j = 0; i + j <= S.length(); j++) {
      string ss;
      for (int y = j; y < j + i; y++) ss += S[y];
      int flag = 1;
      for (int y = 0; y < n; y++) {
        if (Mp[y].find(ss) != Mp[y].end()) {
          if (Mp[y][ss] > b[y] || Mp[y][ss] < a[y]) {
            flag = 0;
          }
        } else if (a[y] > 0) {
          flag = 0;
        }
      }
      if (flag) {
        Se.insert(ss);
      }
    }
  cout << Se.size() << endl;
  cin >> n;
}

```