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 inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
int Pow(int base, int n, int mo) {
  if (n == 0) return 1;
  if (n == 1) return base % mo;
  int tmp = Pow(base, n >> 1, mo);
  tmp = (long long)tmp * tmp % mo;
  if (n & 1) tmp = (long long)tmp * base % mo;
  return tmp;
}
struct node {
  node *nxt[28], *f;
  int l, s[12];
  node() {
    l = 0;
    memset(s, 0, sizeof(s));
    for (int i = (0); i <= (27); ++i) nxt[i] = NULL;
    f = NULL;
  }
} * rt, *las;
int Len = 0;
vector<node *> E[600000];
void insert(char c, int id) {
  Len++;
  node *p = las, *np = new node();
  np->l = p->l + 1;
  E[np->l].push_back(np);
  np->s[id] = 1;
  c -= 'a';
  las = np;
  for (; p && !p->nxt[c]; p = p->f) p->nxt[c] = np;
  if (!p)
    np->f = rt;
  else {
    node *q = p->nxt[c];
    if (q->l == p->l + 1)
      np->f = q;
    else {
      node *r = new node();
      *r = *q;
      memset(r->s, 0, sizeof(r->s));
      r->l = p->l + 1;
      E[r->l].push_back(r);
      np->f = q->f = r;
      for (; p && p->nxt[c] == q; p = p->f) p->nxt[c] = r;
    }
  }
}
int L, M, l[12], r[12], len[12][600000];
char s[60000], buf[60000];
long long ans;
node *p[12][60000];
int main() {
  rt = new node();
  las = rt;
  scanf("%s", s + 1);
  L = strlen(s + 1);
  for (int i = (1); i <= (L); ++i) insert(s[i], 0);
  scanf("%d", &M);
  insert('z' + 1, 11);
  for (int i = (1); i <= (M); ++i) {
    scanf("%s", buf + 1);
    int len = strlen(buf + 1);
    for (int j = (1); j <= (len); ++j) insert(buf[j], i);
    insert('z' + 1, 11);
    scanf("%d%d", &l[i], &r[i]);
  }
  for (int i = (Len); i >= (1); --i)
    for (int j = 0; j < E[i].size(); ++j) {
      node *p = E[i][j];
      if (p->f)
        for (int j = (0); j <= (M); ++j) p->f->s[j] += p->s[j];
      int valid = 1;
      for (int k = (1); k <= (M); ++k)
        if (!(p->s[k] >= l[k] && p->s[k] <= r[k])) valid = 0;
      if (valid && p->s[0]) ans += p->l - p->f->l;
    }
  printf("%lld\n", ans);
  return 0;
}
