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 = 2000 + 10;
int f[N];
int ch[N * 1000][26], sz, ret;
int num, l[11], r[11];
char s[N], p[11][N];
int vis[N], cnt[N], key[N * 1000];
void getfail(char *p, int f[]) {
  int m = strlen(p);
  f[0] = 0;
  f[1] = 0;
  for (int i = 1; i < m; i++) {
    int j = f[i];
    while (j && p[i] != p[j]) j = f[j];
    f[i + 1] = p[i] == p[j] ? j + 1 : 0;
  }
}
void init() {
  memset(ch[0], 0, sizeof(ch[0]));
  memset(key, 0, sizeof(key));
  sz = 1;
}
void insert(int &rt, int k, char *s) {
  int m = strlen(s);
  for (int i = k; i < m; i++) {
    int c = s[i] - 'a';
    if (ch[rt][c] == 0) {
      memset(ch[sz], 0, sizeof(ch[sz]));
      ch[rt][c] = sz;
      sz++;
    }
    rt = ch[rt][c];
  }
  if (key[rt] == 0) {
    ret++;
    key[rt] = 1;
  }
}
int find(char *T, char *p, int f[]) {
  int j = 0;
  int n = strlen(T);
  int m = strlen(p);
  for (int i = 0; i < n; i++) {
    while (j && T[i] != p[j]) j = f[j];
    if (T[i] == p[j]) j++;
    cnt[j]++;
  }
}
void work() {
  char *q = s;
  char ss[N];
  int len = strlen(s);
  for (int i = 0; i < len; i++) {
    getfail(q, f);
    memset(vis, 0, sizeof(vis));
    for (int j = 0; j < num; j++) {
      memset(cnt, 0, sizeof(cnt));
      find(p[j], q, f);
      for (int k = 1; k <= len - i; k++) {
        if (cnt[k] < l[j] || cnt[k] > r[j]) vis[k] = 1;
      }
    }
    int rt = 0, k = 0;
    for (int j = 1; j <= len - i; j++) {
      ss[j - 1] = s[i + j - 1];
      if (vis[j] == 0) {
        ss[j] = '\0';
        insert(rt, k, ss);
        k = j;
      }
    }
    q++;
  }
}
int main() {
  while (~scanf("%s", s)) {
    scanf("%d", &num);
    memset(vis, 0, sizeof(vis));
    init();
    for (int i = 0; i < num; i++) {
      scanf("%s", p[i]);
      scanf("%d%d", &l[i], &r[i]);
    }
    ret = 0;
    work();
    cout << ret << endl;
  }
  return 0;
}
