#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
vector<long long> pwr(50000 + 5);
void solve() {
  string s;
  long long n;
  cin >> s >> n;
  vector<string> v(n);
  vector<pair<long long, long long> > lr(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i] >> lr[i].first >> lr[i].second;
  }
  map<long long, long long> uni, each[n];
  int alp = 27;
  vector<long long> pref(s.size() + 1, 0);
  long long sz = s.size();
  for (long long j = 0; j < sz; j++) {
    pref[j + 1] = (pref[j] * alp) % mod + (s[j] - 'a' + 1);
    pref[j + 1] %= mod;
  }
  for (int j = 1; j <= sz; j++) {
    for (int xx, z = 0; z < j; z++) {
      xx = pref[j] - (pref[z] * pwr[j - z]) % mod;
      xx %= mod;
      if (xx < 0) xx += mod;
      uni[xx]++;
    }
  }
  vector<long long> subs, fre;
  while (uni.size()) {
    subs.push_back(uni.begin()->first);
    fre.push_back(uni.begin()->second);
    uni.erase(uni.begin());
  }
  for (long long i = 0; i < n; i++) {
    vector<long long> pref(v[i].size() + 1, 0);
    long long sz = v[i].size();
    for (long long j = 0; j < sz; j++) {
      pref[j + 1] = (pref[j] * alp) % mod + (v[i][j] - 'a' + 1);
      pref[j + 1] %= mod;
    }
    for (int j = 1; j <= sz; j++) {
      for (int xx, z = 0; z < j; z++) {
        xx = pref[j] - (pref[z] * pwr[j - z]) % mod;
        xx %= mod;
        if (xx < 0) xx += mod;
        each[i][xx]++;
      }
    }
  }
  long long ans = 0;
  long long nos = subs.size();
  for (long long i = 0; i < nos; i++) {
    int cnt = 0;
    for (long long val, j = 0; j < n; j++) {
      val = each[j][subs[i]];
      if (lr[j].first <= val and val <= lr[j].second) cnt++;
    }
    if (cnt == n) ans += 1;
  }
  cout << ans << "\n";
}
int main() {
  pwr[0] = 1;
  for (long long i = 1; i <= 50000 + 4; i++) pwr[i] = (pwr[i - 1] * 27) % mod;
  long long t = 1;
  while (t--) {
    solve();
  }
  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 mod = 1e9 + 7;
vector<long long> pwr(50000 + 5);
void solve() {
  string s;
  long long n;
  cin >> s >> n;
  vector<string> v(n);
  vector<pair<long long, long long> > lr(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i] >> lr[i].first >> lr[i].second;
  }
  map<long long, long long> uni, each[n];
  int alp = 27;
  vector<long long> pref(s.size() + 1, 0);
  long long sz = s.size();
  for (long long j = 0; j < sz; j++) {
    pref[j + 1] = (pref[j] * alp) % mod + (s[j] - 'a' + 1);
    pref[j + 1] %= mod;
  }
  for (int j = 1; j <= sz; j++) {
    for (int xx, z = 0; z < j; z++) {
      xx = pref[j] - (pref[z] * pwr[j - z]) % mod;
      xx %= mod;
      if (xx < 0) xx += mod;
      uni[xx]++;
    }
  }
  vector<long long> subs, fre;
  while (uni.size()) {
    subs.push_back(uni.begin()->first);
    fre.push_back(uni.begin()->second);
    uni.erase(uni.begin());
  }
  for (long long i = 0; i < n; i++) {
    vector<long long> pref(v[i].size() + 1, 0);
    long long sz = v[i].size();
    for (long long j = 0; j < sz; j++) {
      pref[j + 1] = (pref[j] * alp) % mod + (v[i][j] - 'a' + 1);
      pref[j + 1] %= mod;
    }
    for (int j = 1; j <= sz; j++) {
      for (int xx, z = 0; z < j; z++) {
        xx = pref[j] - (pref[z] * pwr[j - z]) % mod;
        xx %= mod;
        if (xx < 0) xx += mod;
        each[i][xx]++;
      }
    }
  }
  long long ans = 0;
  long long nos = subs.size();
  for (long long i = 0; i < nos; i++) {
    int cnt = 0;
    for (long long val, j = 0; j < n; j++) {
      val = each[j][subs[i]];
      if (lr[j].first <= val and val <= lr[j].second) cnt++;
    }
    if (cnt == n) ans += 1;
  }
  cout << ans << "\n";
}
int main() {
  pwr[0] = 1;
  for (long long i = 1; i <= 50000 + 4; i++) pwr[i] = (pwr[i - 1] * 27) % mod;
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```