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 L = 50500, C = 27, N = 11, root = 0, None = -987958798;
const char END = '{';
struct node {
  int s[C], b[N], par, val;
} t[L * 22] = {};
char ch[L * 22] = {};
int n, l = 0, tot = 0, x[N] = {}, y[N] = {}, d[N] = {}, in[L * 22] = {},
       q[L * 22] = {}, head = 0, tail = 0;
long long ans = 0;
void add(int p, int c) {
  int np = ++tot;
  t[np].val = t[p].val + 1;
  while (p != None && t[p].s[c] == 0) t[p].s[c] = np, p = t[p].par;
  if (p == None)
    t[np].par = root;
  else {
    int q = t[p].s[c];
    if (t[p].val + 1 == t[q].val)
      t[np].par = q;
    else {
      int nq = ++tot;
      t[nq] = t[q];
      t[nq].val = t[p].val + 1;
      t[q].par = t[np].par = nq;
      for (int i = p; i != None && t[i].s[c] == q; i = t[i].par) t[i].s[c] = nq;
    }
  }
}
int main() {
  t[root].par = None;
  d[0] = ++l;
  cin >> (ch + d[0]) >> n;
  l += strlen(ch + d[0]);
  for (int i = 1; i <= n; ++i) {
    ch[l] = END;
    d[i] = ++l;
    cin >> (ch + d[i]) >> x[i] >> y[i];
    l += strlen(ch + d[i]);
  }
  ch[l] = END;
  for (int i = 1, p = root; i < l; p = t[p].s[ch[i++] - 'a'])
    add(p, ch[i] - 'a');
  for (int i = 0; i <= n; ++i)
    for (int j = d[i], p = t[root].s[ch[j] - 'a']; ch[j] != END;
         p = t[p].s[ch[++j] - 'a'])
      ++t[p].b[i];
  for (int i = 1; i <= tot; ++i) ++in[t[i].par];
  for (int i = 1; i <= tot; ++i)
    if (!in[i]) q[++tail] = i;
  for (int tmp = 1; tmp <= tot; ++tmp) {
    int s = q[++head];
    bool flag = t[s].b[0];
    for (int i = 1; i <= n; ++i) flag &= x[i] <= t[s].b[i] && t[s].b[i] <= y[i];
    if (flag) ans += t[s].val - t[t[s].par].val;
    for (int i = 0; i <= n; ++i) t[t[s].par].b[i] += t[s].b[i];
    if ((--in[t[s].par]) == 0) q[++tail] = t[s].par;
  }
  cout << ans << endl;
}
