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;
const int N = 2222;
vector<string> t;
vector<int> lo, hi;
int sum[N];
bool ok[N];
int st[N];
int res = 0;
void cnt(string b, string a) {
  string s = b + '&' + a;
  int n = s.length();
  vector<int> z(n, 0);
  for (int i = 1, l = 0, r = 0; i < n; i++) {
    if (i <= r) z[i] = min(z[i - l], r - i + 1);
    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;
    }
  }
  for (int i = 0; i < n; i++) {
    if (i > b.length()) sum[z[i]]++;
  }
}
void work(string a, int start) {
  int n = a.length();
  fill(ok, ok + n + 1, 1);
  for (int i = 0; i < t.size(); i++) {
    fill(sum, sum + n + 1, 0);
    cnt(a, t[i]);
    for (int j = n - 1; j >= 0; j--) sum[j] += sum[j + 1];
    for (int j = 0; j <= n; j++)
      if (sum[j] < lo[i] || sum[j] > hi[i]) ok[j] = false;
  }
  for (int j = start; j <= n; j++)
    if (ok[j]) res++;
}
int main() {
  ios_base::sync_with_stdio(0);
  string sss;
  cin >> sss;
  int q;
  cin >> q;
  t.resize(q);
  lo.resize(q);
  hi.resize(q);
  for (int i = 0; i < q; i++) cin >> t[i] >> lo[i] >> hi[i];
  for (int f = 0; f < sss.length(); f++) {
    string t = sss.substr(f, sss.length() - f);
    string s = t + '$' + sss;
    int n = s.length();
    vector<int> z(n, 0);
    for (int i = 1, l = 0, r = 0; i < n; i++) {
      if (i <= r) z[i] = min(z[i - l], r - i + 1);
      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;
      }
    }
    st[f] = 1;
    for (int i = 0; i < n; i++)
      if (i > t.length() && i - (int)t.length() - 1 < f)
        st[f] = max(st[f], z[i] + 1);
  }
  int n = sss.length();
  for (int i = 0; i < n; i++) {
    string a = sss.substr(i, n - i);
    work(a, st[i]);
  }
  cout << res;
  return 0;
}
