#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2002;
int N;
int d[11];
int u[11];
int hash[11][MAXN][MAXN];
string s;
string q[11];
set<int> S;
map<int, int> H[11];
int main() {
  cin >> s;
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> q[i] >> d[i] >> u[i];
  for (int k = 1; k <= N; k++)
    for (int i = 0; i < ((int)q[k].size()); i++) {
      int h = 0;
      for (int j = i; j < ((int)q[k].size()); j++) {
        h = h * 157 + q[k][j];
        H[k][h]++;
      }
    }
  int res = 0;
  for (int i = 0; i < ((int)s.size()); i++) {
    int h = 0;
    for (int j = i; j < ((int)s.size()); j++) {
      h = h * 157LL + s[j];
      if (S.find(h) == S.end()) {
        S.insert(h);
        bool ok = 1;
        for (int k = 1; k <= N; k++)
          if (H[k][h] < d[k] || H[k][h] > u[k]) {
            ok = 0;
            break;
          }
        res += ok;
      }
    }
  }
  printf("%d\n", res);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MAXN = 2002;
int N;
int d[11];
int u[11];
int hash[11][MAXN][MAXN];
string s;
string q[11];
set<int> S;
map<int, int> H[11];
int main() {
  cin >> s;
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> q[i] >> d[i] >> u[i];
  for (int k = 1; k <= N; k++)
    for (int i = 0; i < ((int)q[k].size()); i++) {
      int h = 0;
      for (int j = i; j < ((int)q[k].size()); j++) {
        h = h * 157 + q[k][j];
        H[k][h]++;
      }
    }
  int res = 0;
  for (int i = 0; i < ((int)s.size()); i++) {
    int h = 0;
    for (int j = i; j < ((int)s.size()); j++) {
      h = h * 157LL + s[j];
      if (S.find(h) == S.end()) {
        S.insert(h);
        bool ok = 1;
        for (int k = 1; k <= N; k++)
          if (H[k][h] < d[k] || H[k][h] > u[k]) {
            ok = 0;
            break;
          }
        res += ok;
      }
    }
  }
  printf("%d\n", res);
  return 0;
}

```