#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:16000000")
const int INF = 1e9;
const double EPS = 1e-8;
const double PI = 2 * acos(0.);
const long long P = 909091;
map<long long, int> M[11];
char s[3000];
char p[3000];
long long h[3000];
long long ppow[100500];
int l[3000], r[3000];
set<long long> S;
void make_hash(char* s) {
  h[0] = 0;
  for (int i = 0; s[i] != 0; ++i) {
    h[i + 1] = h[i] * P + s[i];
  }
}
int get_hash(int l, int r) { return h[r + 1] - h[l] * ppow[r - l + 1]; }
int main() {
  ppow[0] = 1;
  for (int i = 1; i < 100500; ++i) {
    ppow[i] = ppow[i - 1] * P;
  }
  scanf("%s", s);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s%d%d", p, &l[i], &r[i]);
    make_hash(p);
    for (int L = 0; p[L] != 0; ++L) {
      for (int R = L; p[R] != 0; ++R) {
        M[i][get_hash(L, R)]++;
      }
    }
  }
  make_hash(s);
  for (int L = 0; s[L] != 0; ++L) {
    for (int R = L; s[R] != 0; ++R) {
      long long H = get_hash(L, R);
      bool ok = true;
      for (int i = 0; i < n; ++i) {
        if (M[i][H] < l[i] || M[i][H] > r[i]) {
          ok = false;
          break;
        }
      }
      if (ok) S.insert(H);
    }
  }
  printf("%d\n", S.size());
  return 0;
}

### 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;
#pragma comment(linker, "/STACK:16000000")
const int INF = 1e9;
const double EPS = 1e-8;
const double PI = 2 * acos(0.);
const long long P = 909091;
map<long long, int> M[11];
char s[3000];
char p[3000];
long long h[3000];
long long ppow[100500];
int l[3000], r[3000];
set<long long> S;
void make_hash(char* s) {
  h[0] = 0;
  for (int i = 0; s[i] != 0; ++i) {
    h[i + 1] = h[i] * P + s[i];
  }
}
int get_hash(int l, int r) { return h[r + 1] - h[l] * ppow[r - l + 1]; }
int main() {
  ppow[0] = 1;
  for (int i = 1; i < 100500; ++i) {
    ppow[i] = ppow[i - 1] * P;
  }
  scanf("%s", s);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s%d%d", p, &l[i], &r[i]);
    make_hash(p);
    for (int L = 0; p[L] != 0; ++L) {
      for (int R = L; p[R] != 0; ++R) {
        M[i][get_hash(L, R)]++;
      }
    }
  }
  make_hash(s);
  for (int L = 0; s[L] != 0; ++L) {
    for (int R = L; s[R] != 0; ++R) {
      long long H = get_hash(L, R);
      bool ok = true;
      for (int i = 0; i < n; ++i) {
        if (M[i][H] < l[i] || M[i][H] > r[i]) {
          ok = false;
          break;
        }
      }
      if (ok) S.insert(H);
    }
  }
  printf("%d\n", S.size());
  return 0;
}

```