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;
struct tnode {
  bool ok;
  int kv;
  vector<pair<int, tnode *> > link;
  tnode() {
    ok = true;
    kv = 0;
    link.resize(0);
  }
};
typedef tnode *pnode;
const int maxn = 50005;
char s[maxn];
char p[20][maxn];
int len[20];
int l[20], r[20];
int n, m;
int answer;
void add(pnode cur, char *s, int curc, int len) {
  if (curc == len) return;
  int to_find = s[curc] - 'a';
  int wh = -1;
  for (int i = 0; i < (int)cur->link.size(); i++)
    if (cur->link[i].first == to_find) {
      wh = i;
      break;
    }
  if (wh == -1) {
    wh = cur->link.size();
    cur->link.push_back(make_pair(to_find, new tnode));
  }
  add(cur->link[wh].second, s, curc + 1, len);
}
void check(pnode cur, char *s, int curc, int len) {
  cur->kv++;
  if (curc == len) return;
  int to_find = s[curc] - 'a';
  int wh = -1;
  for (int i = 0; i < (int)cur->link.size(); i++)
    if (cur->link[i].first == to_find) {
      wh = i;
      break;
    }
  if (wh == -1) return;
  check(cur->link[wh].second, s, curc + 1, len);
}
void goall(pnode cur, int l, int r) {
  if (cur->kv < l || cur->kv > r) cur->ok = false;
  cur->kv = 0;
  for (int i = 0; i < (int)cur->link.size(); i++)
    goall(cur->link[i].second, l, r);
}
void calc(pnode cur) {
  if (cur->ok) answer++;
  for (int i = 0; i < (int)cur->link.size(); i++) calc(cur->link[i].second);
}
int main() {
  scanf("%s", s);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s%d%d", p[i], &l[i], &r[i]);
    len[i] = strlen(p[i]);
  }
  int m = strlen(s);
  pnode root = new tnode;
  for (int i = 0; i < m; i++) add(root, s, i, m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < len[i]; j++) check(root, p[i], j, len[i]);
    goall(root, l[i], r[i]);
  }
  answer = 0;
  root->ok = false;
  calc(root);
  cout << answer << endl;
  return 0;
}
