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;
int Get() {
  char c;
  while (c = getchar(), (c < '0' || c > '9') && (c != '-'))
    ;
  bool Flag = (c == '-');
  if (Flag) c = getchar();
  int X = 0;
  while (c >= '0' && c <= '9') {
    X = X * 10 + c - 48;
    c = getchar();
  }
  return Flag ? -X : X;
}
void Output(int X) {
  if (X < 0) {
    putchar('-');
    X = -X;
  }
  int Len = 0, Data[10];
  while (X) {
    Data[Len++] = X % 10;
    X /= 10;
  }
  if (!Len) Data[Len++] = 0;
  while (Len--) putchar(Data[Len] + 48);
  putchar('\n');
}
const int Base = 19961021;
long long Hash[11][2001];
long long Pow[2001];
long long GetHash(int P, int X, int Len) {
  return Hash[P][X + Len] - Hash[P][X] * Pow[Len];
}
int main() {
  string S[11];
  cin >> S[0];
  int N = Get();
  static int LB[11], UB[11];
  for (int i = 1; i <= N; i++) cin >> S[i] >> LB[i] >> UB[i];
  for (int i = 0; i <= N; i++) {
    Hash[i][0] = 0;
    for (int j = 0; j < S[i].size(); j++)
      Hash[i][j + 1] = Hash[i][j] * Base + S[i][j];
  }
  Pow[0] = 1;
  for (int i = 0; i < 2000; i++) Pow[i + 1] = Pow[i] * Base;
  int Total = 0;
  static long long Ans[2000000];
  for (int i = 0; i < S[0].size(); i++) {
    static int Sum[11][2001];
    memset(Sum, 0, sizeof(Sum));
    for (int j = 1; j <= N; j++)
      for (int k = 0; k < S[j].size(); k++) {
        int Left = 0, Right = min(S[0].size() - i, S[j].size() - k);
        while (Left < Right) {
          int Mid = (Left + Right + 1) / 2;
          if (GetHash(0, i, Mid) == GetHash(j, k, Mid))
            Left = Mid;
          else
            Right = Mid - 1;
        }
        Sum[j][Left]++;
      }
    for (int j = 1; j <= N; j++)
      for (int k = S[0].size() - i; k >= 1; k--) Sum[j][k - 1] += Sum[j][k];
    for (int j = 1; j <= S[0].size() - i; j++) {
      bool OK = true;
      for (int k = 1; k <= N; k++)
        if (Sum[k][j] < LB[k] || Sum[k][j] > UB[k]) OK = false;
      if (OK) Ans[Total++] = GetHash(0, i, j);
    }
  }
  sort(Ans, Ans + Total);
  int Count = 0;
  for (int i = 0; i < Total; i++)
    if (!i || Ans[i] != Ans[i - 1]) Count++;
  Output(Count);
  return 0;
}
