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;
#pragma comment(linker, "/STACK:255000000")
bool firstout = 1;
template <class T>
T &minn(T &a, T b) {
  if (b < a) a = b;
  return a;
}
template <class T>
T &maxx(T &a, T b) {
  if (a < b) a = b;
  return a;
}
int &madd(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
int &msub(int &a, int b) {
  a -= b;
  if (a < 0) a += 1000000007;
  return a;
}
int &mmult(int &a, int b) { return a = (long long)a * b % 1000000007; }
int mdiv(long long a, long long b, long long m) {
  a = (a % m + m) % m;
  b = (b % m + m) % m;
  if (a % b == 0) return a / b;
  return (a + m * mdiv(-a, m, b)) / b;
}
int n, m, q;
char S[51234];
char A[11][51234];
int L[11], R[11];
long long P[51234];
long long HS[51234];
void calc(char *s, int m, vector<pair<long long, int> > &res) {
  int n = strlen(s);
  if (m > n) {
    res.clear();
    return;
  }
  int i, j;
  long long h = 0;
  for (i = (0); i < (n); ++i) {
    h = 137 * h + s[i];
    if (i >= m) h -= P[m] * s[i - m];
    if (i >= m - 1) HS[i - m + 1] = h;
  }
  int l = n - m + 1;
  sort(HS, HS + l);
  for (i = 0; i < l; i = j) {
    for (j = i; j < l && HS[i] == HS[j]; ++j)
      ;
    res.push_back(make_pair(HS[i], j - i));
  }
}
int main() {
  int i, j, k;
  char c;
  int a, d;
  P[0] = 1;
  for (i = (1); i < (51234); ++i) P[i] = 137 * P[i - 1];
  int ts;
  for (ts = 1; scanf("%"
                     "s",
                     &(S)) > 0;
       ++ts) {
    scanf(
        "%"
        "d",
        &(m));
    for (i = (0); i < (m); ++i)
      scanf(
          "%"
          "s",
          &(A[i])),
          scanf(
              "%"
              "d",
              &(L[i])),
          scanf(
              "%"
              "d",
              &(R[i]));
    n = strlen(S);
    int res = 0;
    int x;
    for (x = (1); x < (n + 1); ++x) {
      vector<pair<long long, int> > h[11];
      for (j = (0); j < (m); ++j) calc(A[j], x, h[j]);
      vector<pair<long long, int> > hh;
      calc(S, x, hh);
      for (i = (0); i < (hh.size()); ++i) {
        long long a = hh[i].first;
        bool g = 1;
        for (j = (0); j < (m); ++j) {
          k = lower_bound((h[j]).begin(), (h[j]).end(), make_pair(a, -1)) -
              h[j].begin();
          if (k < h[j].size() && h[j][k].first == a) {
            if (!(L[j] <= h[j][k].second && h[j][k].second <= R[j])) g = 0;
          } else {
            if (L[j] > 0) g = 0;
          }
        }
        res += g;
      }
    }
    printf(
        "%"
        "d",
        (res));
    printf("\n"), firstout = 1;
  }
  return 0;
}
