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 long long maxn = 2000 + 10, MOD = 29339742923846321ll;
string s, p[20];
long long po[maxn];
long long l[20], r[20], h[maxn][maxn], h1[maxn][maxn], n, ans1 = 0;
vector<long long> ped[maxn], ans;
int main() {
  ios_base::sync_with_stdio(false);
  po[0] = 1;
  for (long long i = 1; i < maxn; i++) po[i] = (po[i - 1] * 26) % MOD;
  cin >> s;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> p[i] >> l[i] >> r[i];
  for (long long i = 0; i < s.size(); i++) {
    h[i][i] = (po[0] * (s[i] - 'a' + 1)) % MOD;
    for (long long j = i + 1; j < s.size(); j++)
      h[i][j] = (h[i][j - 1] + po[j - i] * (s[j] - 'a' + 1) % MOD) % MOD;
  }
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < p[i].size(); j++) {
      h1[j][j] = (po[0] * (p[i][j] - 'a' + 1)) % MOD;
      ped[i].push_back(h1[j][j]);
      for (long long k = j + 1; k < p[i].size(); k++) {
        h1[j][k] = (h1[j][k - 1] + po[k - j] * (p[i][k] - 'a' + 1) % MOD) % MOD;
        ped[i].push_back(h1[j][k]);
      }
    }
  for (int i = 0; i < n; i++) sort(ped[i].begin(), ped[i].end());
  for (long long i = 0; i < s.size(); i++)
    for (long long j = i; j < s.size(); j++) ans.push_back(h[i][j]);
  sort(ans.begin(), ans.end());
  int N = int(unique(ans.begin(), ans.end()) - ans.begin());
  for (int i = 0; i < N; i++) {
    bool sw = true;
    for (long long k = 0; k < n; k++) {
      long long hash = ans[i];
      long long ped1 = upper_bound(ped[k].begin(), ped[k].end(), ans[i]) -
                       lower_bound(ped[k].begin(), ped[k].end(), ans[i]);
      if (ped1 >= l[k] && ped1 <= r[k]) {
      } else {
        sw = false;
        break;
      }
    }
    if (sw) {
      ans1++;
    }
  }
  cout << ans1 << endl;
  return 0;
}
