#include <bits/stdc++.h>
using namespace std;
char s[2002];
char A[12][2002];
int L[12], R[2002], tam[12][2002];
unsigned int HASH[10][2002][2002];
int main() {
  int n;
  while (scanf("%s", s) == 1) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%s %d %d", A[i], &L[i], &R[i]);
    int l = strlen(s);
    set<pair<int, unsigned int> > S;
    for (int i = 0; i < l; i++) {
      unsigned int H = 0;
      for (int j = i; j < l; j++) {
        H = H * 27 + (s[j] - 'a' + 1);
        S.insert(make_pair(j - i + 1, H));
      }
    }
    memset(tam, 0, sizeof(tam));
    for (int ii = 0; ii < n; ii++) {
      int t = strlen(A[ii]);
      for (int i = 0; i < t; i++) {
        unsigned int H = 0;
        for (int j = i; j < t; j++) {
          H = H * 27 + (A[ii][j] - 'a' + 1);
          HASH[ii][j - i + 1][tam[ii][j - i + 1]++] = H;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      int l = strlen(A[i]);
      for (int j = 0; j < l; j++) sort(HASH[i][j], HASH[i][j] + tam[i][j]);
    }
    vector<pair<int, unsigned int> > v(S.begin(), S.end());
    int size = v.size();
    int ans = 0;
    for (int i = 0; i < size; i++) {
      bool ok = 1;
      int t = v[i].first;
      unsigned int find = v[i].second;
      for (int j = 0; ok && j < n; j++) {
        long long g = upper_bound(HASH[j][t], HASH[j][t] + tam[j][t], find) -
                      lower_bound(HASH[j][t], HASH[j][t] + tam[j][t], find);
        if (L[j] <= g && g <= R[j]) continue;
        ok = 0;
      }
      if (ok) {
        ans++;
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
char s[2002];
char A[12][2002];
int L[12], R[2002], tam[12][2002];
unsigned int HASH[10][2002][2002];
int main() {
  int n;
  while (scanf("%s", s) == 1) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%s %d %d", A[i], &L[i], &R[i]);
    int l = strlen(s);
    set<pair<int, unsigned int> > S;
    for (int i = 0; i < l; i++) {
      unsigned int H = 0;
      for (int j = i; j < l; j++) {
        H = H * 27 + (s[j] - 'a' + 1);
        S.insert(make_pair(j - i + 1, H));
      }
    }
    memset(tam, 0, sizeof(tam));
    for (int ii = 0; ii < n; ii++) {
      int t = strlen(A[ii]);
      for (int i = 0; i < t; i++) {
        unsigned int H = 0;
        for (int j = i; j < t; j++) {
          H = H * 27 + (A[ii][j] - 'a' + 1);
          HASH[ii][j - i + 1][tam[ii][j - i + 1]++] = H;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      int l = strlen(A[i]);
      for (int j = 0; j < l; j++) sort(HASH[i][j], HASH[i][j] + tam[i][j]);
    }
    vector<pair<int, unsigned int> > v(S.begin(), S.end());
    int size = v.size();
    int ans = 0;
    for (int i = 0; i < size; i++) {
      bool ok = 1;
      int t = v[i].first;
      unsigned int find = v[i].second;
      for (int j = 0; ok && j < n; j++) {
        long long g = upper_bound(HASH[j][t], HASH[j][t] + tam[j][t], find) -
                      lower_bound(HASH[j][t], HASH[j][t] + tam[j][t], find);
        if (L[j] <= g && g <= R[j]) continue;
        ok = 0;
      }
      if (ok) {
        ans++;
      }
    }
    cout << ans << endl;
  }
}

```