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 MAXN = 15;
const int MAXL = 50005;
const int MAXP = 1200005;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
struct SuffixAutomaton {
  struct Node {
    int child[26];
    int size[MAXN];
    int father, depth;
  } a[MAXP];
  char s[MAXL];
  vector<int> b[MAXP];
  int l[MAXL], r[MAXL];
  int n, len, root, last, size;
  long long ans;
  int newnode(int dep) {
    a[size].depth = dep;
    return size++;
  }
  void extend(int ch, int col) {
    int p = last;
    if (!a[p].child[ch]) {
      int np = newnode(a[p].depth + 1);
      while (!a[p].child[ch]) {
        a[p].child[ch] = np;
        p = a[p].father;
      }
      if (a[p].child[ch] == np)
        last = np;
      else {
        int q = a[p].child[ch];
        if (a[p].depth + 1 == a[q].depth)
          a[np].father = q, last = np;
        else {
          int nq = newnode(a[p].depth + 1);
          memcpy(a[nq].child, a[q].child, sizeof(a[q].child));
          a[nq].father = a[q].father;
          a[q].father = a[np].father = nq;
          while (a[p].child[ch] == q) {
            a[p].child[ch] = nq;
            p = a[p].father;
          }
          last = np;
        }
      }
    } else {
      int q = a[p].child[ch];
      if (a[p].depth + 1 == a[q].depth)
        last = q;
      else {
        int np = newnode(a[p].depth + 1);
        memcpy(a[np].child, a[q].child, sizeof(a[q].child));
        a[np].father = a[q].father;
        a[q].father = np;
        while (a[p].child[ch] == q) {
          a[p].child[ch] = np;
          p = a[p].father;
        }
        last = np;
      }
    }
    a[last].size[col]++;
  }
  void init() {
    size = 0;
    last = root = newnode(0);
    scanf("\n%s", s + 1);
    len = strlen(s + 1);
    for (int i = 1; i <= len; i++) extend(s[i] - 'a', 0);
    read(n);
    for (int i = 1; i <= n; i++) {
      last = root;
      scanf("\n%s", s + 1);
      len = strlen(s + 1);
      for (int j = 1; j <= len; j++) extend(s[j] - 'a', i);
      read(l[i]), read(r[i]);
    }
  }
  void work(int pos) {
    for (unsigned i = 0; i < b[pos].size(); i++) {
      int tmp = b[pos][i];
      work(tmp);
      for (int j = 0; j <= n; j++) a[pos].size[j] += a[tmp].size[j];
    }
    if (pos) {
      bool flg = a[pos].size[0] != 0;
      for (int j = 1; j <= n; j++)
        flg &= a[pos].size[j] >= l[j] && a[pos].size[j] <= r[j];
      ans += flg * (a[pos].depth - a[a[pos].father].depth);
    }
  }
  void calc() {
    for (int i = 1; i < size; i++) b[a[i].father].push_back(i);
    work(0);
    writeln(ans);
  }
} SAM;
int main() {
  SAM.init();
  SAM.calc();
  return 0;
}
