#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;
const ull mod = 1145141919810;
const int maxn = 220;
int n, m, na, nb;
string t;
set<string> sub;
string s[maxn];
int l[maxn], r[maxn];
int pre[maxn];
void getpre(const char* b, int* pre) {
  int j, k;
  pre[0] = -1;
  j = 0;
  k = -1;
  while (j < nb) {
    if (k == -1 || b[j] == b[k]) {
      j++;
      k++;
      if (b[j] != b[k])
        pre[j] = k;
      else
        pre[j] = pre[k];
    } else
      k = pre[k];
  }
}
int kmp(const char* a, const char* b) {
  int ans = 0;
  int i = 0;
  int j = 0;
  while (i < na) {
    if (j == -1 || a[i] == b[j]) {
      i++;
      j++;
    } else
      j = pre[j];
    if (j == nb) ans++;
  }
  return ans;
}
signed main() {
  while (cin >> t) {
    n = t.length();
    sub.clear();
    for (int i = 0; i < n; i++) {
      for (int j = i; j < n; j++) {
        sub.emplace(t.substr(i, j - i + 1));
      }
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
      cin >> s[i] >> l[i] >> r[i];
    }
    if (m == 0) {
      cout << sub.size() << endl;
      continue;
    }
    int ret = 0;
    for (auto p : sub) {
      int tmp = 0;
      nb = p.length();
      getpre(p.c_str(), pre);
      for (int i = 0; i < m; i++) {
        na = s[i].length();
        int tm = kmp(s[i].c_str(), p.c_str());
        if (tm >= l[i] && tm <= r[i]) tmp++;
      }
      if (tmp == m) ret++;
    }
    cout << ret << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
using ull = unsigned long long;
const ull mod = 1145141919810;
const int maxn = 220;
int n, m, na, nb;
string t;
set<string> sub;
string s[maxn];
int l[maxn], r[maxn];
int pre[maxn];
void getpre(const char* b, int* pre) {
  int j, k;
  pre[0] = -1;
  j = 0;
  k = -1;
  while (j < nb) {
    if (k == -1 || b[j] == b[k]) {
      j++;
      k++;
      if (b[j] != b[k])
        pre[j] = k;
      else
        pre[j] = pre[k];
    } else
      k = pre[k];
  }
}
int kmp(const char* a, const char* b) {
  int ans = 0;
  int i = 0;
  int j = 0;
  while (i < na) {
    if (j == -1 || a[i] == b[j]) {
      i++;
      j++;
    } else
      j = pre[j];
    if (j == nb) ans++;
  }
  return ans;
}
signed main() {
  while (cin >> t) {
    n = t.length();
    sub.clear();
    for (int i = 0; i < n; i++) {
      for (int j = i; j < n; j++) {
        sub.emplace(t.substr(i, j - i + 1));
      }
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
      cin >> s[i] >> l[i] >> r[i];
    }
    if (m == 0) {
      cout << sub.size() << endl;
      continue;
    }
    int ret = 0;
    for (auto p : sub) {
      int tmp = 0;
      nb = p.length();
      getpre(p.c_str(), pre);
      for (int i = 0; i < m; i++) {
        na = s[i].length();
        int tm = kmp(s[i].c_str(), p.c_str());
        if (tm >= l[i] && tm <= r[i]) tmp++;
      }
      if (tmp == m) ret++;
    }
    cout << ret << endl;
  }
  return 0;
}

```