#include <bits/stdc++.h>
using namespace std;
string p[10];
int l[10], r[10];
bool goodSubstring[2000][2000];
int matches[2][2000];
pair<int, int> hashes[2000][2000];
const int MOD1 = 999999991;
const int A1 = 21589;
const int MOD2 = 1000000009;
const int A2 = 253;
pair<int, int> nextHash(pair<int, int> hash, char c) {
  hash.first = (1LL * hash.first * A1 + c) % MOD1;
  hash.second = (1LL * hash.second * A2 + c) % MOD2;
  return hash;
}
vector<int> prefix(string s) {
  vector<int> p(s.length() + 1);
  int i, k = 0;
  p[1] = 0;
  for (i = 2; i <= s.length(); i++) {
    while (k && s[k] != s[i - 1]) k = p[k];
    if (s[k] == s[i - 1]) ++k;
    p[i] = k;
  }
  return p;
}
int count_occurences(string &s, string &t) {
  int i, k = 0;
  vector<int> p = prefix(t + string("X") + s);
  int ret = 0;
  for (i = t.length() + 1; i < p.size(); i++)
    if (p[i] == t.length()) ++ret;
  return ret;
}
bool okay(string &t, string &s, int ch, int x) {
  int cnt = count_occurences(s, t);
  if (ch == 0 && cnt < x || ch == 1 && cnt <= x) return false;
  return true;
}
int main() {
  string s;
  int i, j, k, n, m;
  cin >> s;
  cin >> n;
  for (i = 0; i < n; i++) cin >> p[i] >> l[i] >> r[i];
  for (i = 0; i < s.length(); i++)
    for (j = i; j < s.length(); j++) goodSubstring[i][j] = true;
  for (k = 0; k < n; k++) {
    j = 0;
    for (i = 0; i < s.length(); i++) {
      if (j == s.length()) {
        matches[0][i] = j;
        continue;
      }
      string t;
      int q = i;
      while (q <= j) t += s[q++];
      while (j < s.length() && okay(t, p[k], 0, l[k])) {
        ++j;
        if (j == s.length()) break;
        t += s[j];
      }
      matches[0][i] = j;
    }
    j = 0;
    for (i = 0; i < s.length(); i++) {
      if (j == s.length()) {
        matches[0][i] = j;
        continue;
      }
      string t;
      int q = i;
      while (q <= j) t += s[q++];
      while (j < s.length() && okay(t, p[k], 1, r[k])) {
        ++j;
        if (j == s.length()) break;
        t += s[j];
      }
      matches[1][i] = j;
    }
    for (i = 0; i < s.length(); i++) {
      pair<int, int> hash;
      for (j = i; j < s.length(); j++) {
        hash = nextHash(hash, s[j]);
        if (!(matches[0][i] > j && matches[1][i] <= j))
          goodSubstring[i][j] = false;
      }
    }
  }
  for (i = 0; i < s.length(); i++) {
    pair<int, int> hash;
    for (j = i; j < s.length(); j++) {
      hash = nextHash(hash, s[j]);
      hashes[i][j] = hash;
    }
  }
  int ans = 0;
  for (i = 0; i < s.length(); i++) {
    set<pair<int, int> > good;
    for (j = 0; j + i < s.length(); j++)
      if (goodSubstring[j][j + i]) good.insert(hashes[j][j + i]);
    ans += good.size();
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
string p[10];
int l[10], r[10];
bool goodSubstring[2000][2000];
int matches[2][2000];
pair<int, int> hashes[2000][2000];
const int MOD1 = 999999991;
const int A1 = 21589;
const int MOD2 = 1000000009;
const int A2 = 253;
pair<int, int> nextHash(pair<int, int> hash, char c) {
  hash.first = (1LL * hash.first * A1 + c) % MOD1;
  hash.second = (1LL * hash.second * A2 + c) % MOD2;
  return hash;
}
vector<int> prefix(string s) {
  vector<int> p(s.length() + 1);
  int i, k = 0;
  p[1] = 0;
  for (i = 2; i <= s.length(); i++) {
    while (k && s[k] != s[i - 1]) k = p[k];
    if (s[k] == s[i - 1]) ++k;
    p[i] = k;
  }
  return p;
}
int count_occurences(string &s, string &t) {
  int i, k = 0;
  vector<int> p = prefix(t + string("X") + s);
  int ret = 0;
  for (i = t.length() + 1; i < p.size(); i++)
    if (p[i] == t.length()) ++ret;
  return ret;
}
bool okay(string &t, string &s, int ch, int x) {
  int cnt = count_occurences(s, t);
  if (ch == 0 && cnt < x || ch == 1 && cnt <= x) return false;
  return true;
}
int main() {
  string s;
  int i, j, k, n, m;
  cin >> s;
  cin >> n;
  for (i = 0; i < n; i++) cin >> p[i] >> l[i] >> r[i];
  for (i = 0; i < s.length(); i++)
    for (j = i; j < s.length(); j++) goodSubstring[i][j] = true;
  for (k = 0; k < n; k++) {
    j = 0;
    for (i = 0; i < s.length(); i++) {
      if (j == s.length()) {
        matches[0][i] = j;
        continue;
      }
      string t;
      int q = i;
      while (q <= j) t += s[q++];
      while (j < s.length() && okay(t, p[k], 0, l[k])) {
        ++j;
        if (j == s.length()) break;
        t += s[j];
      }
      matches[0][i] = j;
    }
    j = 0;
    for (i = 0; i < s.length(); i++) {
      if (j == s.length()) {
        matches[0][i] = j;
        continue;
      }
      string t;
      int q = i;
      while (q <= j) t += s[q++];
      while (j < s.length() && okay(t, p[k], 1, r[k])) {
        ++j;
        if (j == s.length()) break;
        t += s[j];
      }
      matches[1][i] = j;
    }
    for (i = 0; i < s.length(); i++) {
      pair<int, int> hash;
      for (j = i; j < s.length(); j++) {
        hash = nextHash(hash, s[j]);
        if (!(matches[0][i] > j && matches[1][i] <= j))
          goodSubstring[i][j] = false;
      }
    }
  }
  for (i = 0; i < s.length(); i++) {
    pair<int, int> hash;
    for (j = i; j < s.length(); j++) {
      hash = nextHash(hash, s[j]);
      hashes[i][j] = hash;
    }
  }
  int ans = 0;
  for (i = 0; i < s.length(); i++) {
    set<pair<int, int> > good;
    for (j = 0; j + i < s.length(); j++)
      if (goodSubstring[j][j + i]) good.insert(hashes[j][j + i]);
    ans += good.size();
  }
  cout << ans << endl;
  return 0;
}

```