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(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
struct node {
  int ch[26], fa, len;
} S[1000005];
int K, smt, sls, sz[1000005][11], L[1000005], R[1000005];
struct edge {
  int to, nxt;
} e[1000005];
int n, et, head[1000005];
char ch[1000005];
long long res = 0;
inline void adde(int x, int y) { e[++et] = (edge){y, head[x]}, head[x] = et; }
inline void dfs(int x) {
  for (int i = head[x]; i; i = e[i].nxt) {
    dfs(e[i].to);
    for (int k = 0; k <= K; k++) sz[x][k] += sz[e[i].to][k];
  }
}
inline char chk(int x) {
  for (int i = 1; i <= K; i++)
    if (sz[x][i] < L[i] || sz[x][i] > R[i]) return 0;
  return 1;
}
inline void saminit() { smt = sls = 1; }
inline void samins(int c) {
  int p = sls, np = sls = ++smt;
  S[np].len = S[p].len + 1;
  for (; p && !S[p].ch[c]; p = S[p].fa) S[p].ch[c] = np;
  int q = S[p].ch[c];
  if (!p) return S[np].fa = 1, void();
  if (S[p].len + 1 == S[q].len) return S[np].fa = q, void();
  int nq = ++smt;
  S[nq] = S[q], S[nq].len = S[p].len + 1, S[q].fa = S[np].fa = nq;
  for (; p && S[p].ch[c] == q; p = S[p].fa) S[p].ch[c] = nq;
}
int main() {
  scanf("%s", ch + 1), saminit(), read(K), n = strlen(ch + 1);
  for (int i = 1; i <= n; i++) samins(ch[i] - 'a'), sz[sls][0] = 1;
  for (int k = 1; k <= K; k++) {
    scanf("%s", ch + 1), n = strlen(ch + 1), read(L[k]), read(R[k]);
    sls = 1;
    for (int i = 1; i <= n; i++) samins(ch[i] - 'a'), sz[sls][k] = 1;
  }
  for (int i = 2; i <= smt; i++) adde(S[i].fa, i);
  dfs(1);
  for (int i = 2; i <= smt; i++)
    if (sz[i][0] && chk(i)) res += S[i].len - S[S[i].fa].len;
  return printf("%lld\n", res), 0;
}
