#include <bits/stdc++.h>
using namespace std;
const int N = 202;
const int mod = 1e9 + 7, b = 33;
long long pot[N];
map<int, int> ha[N];
string str;
vector<int> h;
void prep(string &str, vector<int> &hh) {
  hh = vector<int>(str.size() + 1, 0);
  for (int i = 0; i < str.size(); i++) {
    hh[i + 1] = 1LL * hh[i] * b % mod;
    hh[i + 1] += str[i] - 'a' + 1;
    hh[i + 1] %= mod;
  }
}
int getH(vector<int> &hh, int l, int r) {
  return ((1LL * hh[r] - 1LL * hh[l - 1] * pot[r - l + 1]) % mod + mod) % mod;
}
void pr(string &str, map<int, int> &hh) {
  vector<int> hx;
  prep(str, hx);
  for (int l = 1; l <= str.size(); l++)
    for (int r = l; r <= str.size(); r++) {
      hh[getH(hx, l, r)]++;
    }
}
int main(void) {
  ios::sync_with_stdio(false);
  pot[0] = 1;
  for (int i = 1; i < N; i++) pot[i] = pot[i - 1] * b % mod;
  cin >> str;
  prep(str, h);
  int m;
  cin >> m;
  vector<pair<int, int> > qs;
  for (int i = 0; i < m; i++) {
    string aux;
    cin >> aux;
    int l, r;
    cin >> l >> r;
    qs.push_back(pair<int, int>(l, r));
    pr(aux, ha[i]);
  }
  int ans = 0;
  set<int> seen;
  for (int l = 0; l < str.size(); l++)
    for (int r = l; r < str.size(); r++) {
      int hx = 0;
      for (int i = l; i <= r; i++) {
        hx = 1LL * hx * b % mod;
        hx += str[i] - 'a' + 1;
        hx %= mod;
      }
      if (seen.count(hx)) continue;
      seen.insert(hx);
      bool ok = 1;
      for (int i = 0; i < m; i++) {
        int rr = ha[i][hx];
        ok &= rr >= qs[i].first && rr <= qs[i].second;
      }
      ans += ok;
    }
  cout << ans << endl;
  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 N = 202;
const int mod = 1e9 + 7, b = 33;
long long pot[N];
map<int, int> ha[N];
string str;
vector<int> h;
void prep(string &str, vector<int> &hh) {
  hh = vector<int>(str.size() + 1, 0);
  for (int i = 0; i < str.size(); i++) {
    hh[i + 1] = 1LL * hh[i] * b % mod;
    hh[i + 1] += str[i] - 'a' + 1;
    hh[i + 1] %= mod;
  }
}
int getH(vector<int> &hh, int l, int r) {
  return ((1LL * hh[r] - 1LL * hh[l - 1] * pot[r - l + 1]) % mod + mod) % mod;
}
void pr(string &str, map<int, int> &hh) {
  vector<int> hx;
  prep(str, hx);
  for (int l = 1; l <= str.size(); l++)
    for (int r = l; r <= str.size(); r++) {
      hh[getH(hx, l, r)]++;
    }
}
int main(void) {
  ios::sync_with_stdio(false);
  pot[0] = 1;
  for (int i = 1; i < N; i++) pot[i] = pot[i - 1] * b % mod;
  cin >> str;
  prep(str, h);
  int m;
  cin >> m;
  vector<pair<int, int> > qs;
  for (int i = 0; i < m; i++) {
    string aux;
    cin >> aux;
    int l, r;
    cin >> l >> r;
    qs.push_back(pair<int, int>(l, r));
    pr(aux, ha[i]);
  }
  int ans = 0;
  set<int> seen;
  for (int l = 0; l < str.size(); l++)
    for (int r = l; r < str.size(); r++) {
      int hx = 0;
      for (int i = l; i <= r; i++) {
        hx = 1LL * hx * b % mod;
        hx += str[i] - 'a' + 1;
        hx %= mod;
      }
      if (seen.count(hx)) continue;
      seen.insert(hx);
      bool ok = 1;
      for (int i = 0; i < m; i++) {
        int rr = ha[i][hx];
        ok &= rr >= qs[i].first && rr <= qs[i].second;
      }
      ans += ok;
    }
  cout << ans << endl;
  return 0;
}

```