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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
int n, lenth, las = 1, root = 1, tot = 1;
long long ans;
int ch[1000010][26], fa[1000010], len[1000010], siz[1000010][15], l[1000010],
    r[1000010];
char s[1000010], str[1000010];
struct edge {
  int to, nxt;
} e[1000010];
int head[1000010], edge_cnt;
void add(int from, int to) {
  e[++edge_cnt] = (edge){to, head[from]};
  head[from] = edge_cnt;
}
void insert(int c, int id) {
  int p = las, np = las = ++tot;
  len[np] = len[p] + 1, siz[np][id] = 1;
  while (p && !ch[p][c]) ch[p][c] = np, p = fa[p];
  if (!p)
    fa[np] = root;
  else {
    int q = ch[p][c];
    if (len[q] == len[p] + 1)
      fa[np] = q;
    else {
      int nq = ++tot;
      memcpy(ch[nq], ch[q], sizeof(ch[q]));
      len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
      while (ch[p][c] == q) ch[p][c] = nq, p = fa[p];
    }
  }
}
void dfs(int x) {
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    dfs(y);
    for (int j = 0; j <= n; ++j) siz[x][j] += siz[y][j];
  }
}
void calc(int x) {
  if (!siz[x][0]) return;
  for (int i = 1; i <= n; ++i)
    if (siz[x][i] < l[i] || siz[x][i] > r[i]) return;
  ans += len[x] - len[fa[x]];
}
int main() {
  scanf("%s", str + 1), lenth = strlen(str + 1);
  for (int i = 1; i <= lenth; ++i) insert(str[i] - 'a', 0);
  read(n);
  for (int i = 1; i <= n; ++i) {
    las = 1, scanf("%s", s + 1), lenth = strlen(s + 1), read(l[i]), read(r[i]);
    for (int j = 1; j <= lenth; ++j) insert(s[j] - 'a', i);
  }
  for (int i = 2; i <= tot; ++i) add(fa[i], i);
  dfs(root);
  for (int i = 2; i <= tot; ++i) calc(i);
  printf("%lld", ans);
  return 0;
}
