#include <bits/stdc++.h>
using namespace std;
struct KMP {
  string P;
  int nP;
  vector<int> B;
  KMP() {}
  KMP(string s) {
    P = s;
    nP = P.size();
    init();
  }
  void init() {
    B = vector<int>(nP + 1, -1);
    for (int i = 0, j = -1; i < nP; i++) {
      while (j != -1 && P[i] != P[j]) j = B[j];
      j++;
      B[i + 1] = j;
    }
  }
  int match(string &T) {
    int nT = T.size();
    int matches = 0;
    for (int i = 0, j = 0; i < nT; i++) {
      while (j != -1 && T[i] != P[j]) j = B[j];
      j++;
      if (j == nP) matches++, j = B[j];
    }
    return matches;
  }
  bool is_periodic(int pos) {
    int m = pos + 1, b = B[m];
    if (2 * b - m < 0) return 0;
    if ((2 * b - m) % (m - b) == 0) return 1;
    return 0;
  }
  int period(int pos) {
    int m = pos + 1, b = B[m];
    return m - b;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  while (cin >> s) {
    int m;
    cin >> m;
    int Y[m], X[m];
    string t[m];
    for (int i = 0; i < (m); i++) cin >> t[i] >> X[i] >> Y[i];
    int ns = s.size();
    int ans = 0;
    set<string> S;
    for (int i = 0; i < ns; ++i)
      for (int j = i; j < ns; ++j) {
        string temp = s.substr(i, j - i + 1);
        int cnt = 0;
        KMP kmp = KMP(temp);
        for (int k = 0; k < (m); k++) {
          int u = kmp.match(t[k]);
          if (u >= X[k] && u <= Y[k]) cnt++;
        }
        if (cnt == m && !S.count(temp)) ans++;
        S.insert(temp);
      }
    cout << ans << endl;
  }
}

### Prompt

Your challenge is to write 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;
struct KMP {
  string P;
  int nP;
  vector<int> B;
  KMP() {}
  KMP(string s) {
    P = s;
    nP = P.size();
    init();
  }
  void init() {
    B = vector<int>(nP + 1, -1);
    for (int i = 0, j = -1; i < nP; i++) {
      while (j != -1 && P[i] != P[j]) j = B[j];
      j++;
      B[i + 1] = j;
    }
  }
  int match(string &T) {
    int nT = T.size();
    int matches = 0;
    for (int i = 0, j = 0; i < nT; i++) {
      while (j != -1 && T[i] != P[j]) j = B[j];
      j++;
      if (j == nP) matches++, j = B[j];
    }
    return matches;
  }
  bool is_periodic(int pos) {
    int m = pos + 1, b = B[m];
    if (2 * b - m < 0) return 0;
    if ((2 * b - m) % (m - b) == 0) return 1;
    return 0;
  }
  int period(int pos) {
    int m = pos + 1, b = B[m];
    return m - b;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  while (cin >> s) {
    int m;
    cin >> m;
    int Y[m], X[m];
    string t[m];
    for (int i = 0; i < (m); i++) cin >> t[i] >> X[i] >> Y[i];
    int ns = s.size();
    int ans = 0;
    set<string> S;
    for (int i = 0; i < ns; ++i)
      for (int j = i; j < ns; ++j) {
        string temp = s.substr(i, j - i + 1);
        int cnt = 0;
        KMP kmp = KMP(temp);
        for (int k = 0; k < (m); k++) {
          int u = kmp.match(t[k]);
          if (u >= X[k] && u <= Y[k]) cnt++;
        }
        if (cnt == m && !S.count(temp)) ans++;
        S.insert(temp);
      }
    cout << ans << endl;
  }
}

```