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.
#include <bits/stdc++.h>
using namespace std;
vector<int> z_function(const string& s) {
  int n = (int)s.length();
  vector<int> z(n);
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
long long h1 = 57;
long long h2 = 213;
void solve() {
  string s;
  cin >> s;
  int n;
  cin >> n;
  vector<vector<int> > occur(n);
  vector<string> ps;
  vector<int> L, R;
  for (int i = 0; i < int(n); ++i) {
    string p;
    cin >> p;
    ps.push_back(p);
    int l, r;
    cin >> l >> r;
    L.push_back(l);
    R.push_back(r);
  }
  set<pair<long long, long long> > used;
  int ans = 0;
  for (int l = 0; l < int((s).size()); ++l) {
    vector<vector<int> > zfs(n);
    for (int i = 0; i < int(n); ++i) {
      zfs[i] = z_function(s.substr(l, s.size()) + "$" + ps[i]);
      zfs[i].erase(zfs[i].begin(), zfs[i].begin() + s.size() - l + 1);
      sort((zfs[i]).begin(), (zfs[i]).end());
    }
    vector<int> pos(n);
    long long hash1 = 0, hash2 = 0;
    for (int r = l + 1; r <= s.size(); ++r) {
      hash1 = hash1 * h1 + s[r - 1];
      hash2 = hash2 * h2 + s[r - 1];
      auto elem = make_pair(hash1, hash2);
      if (used.count(elem)) continue;
      used.insert(elem);
      bool ok = true;
      for (int i = 0; i < int(n); ++i) {
        while (pos[i] < zfs[i].size() && zfs[i][pos[i]] < r - l) {
          ++pos[i];
        }
        int cnt = zfs[i].size() - pos[i];
        if (cnt < L[i] || cnt > R[i]) {
          ok = false;
          break;
        }
      }
      if (ok) {
        ++ans;
      }
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}
