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;
struct tnode {
  int value, head;
};
typedef tnode* pnode;
pnode root[20], a[8000002];
string d[20];
int trai, phai, n, ans, edge, l[20], r[20], trc[8000002];
pnode st[4000001][2];
char w[8000002];
void build(int id) {
  int i, j, we;
  pnode p, q, v;
  root[id] = new tnode;
  (*root[id]).head = 0;
  for (i = 0; i <= int(d[id].length()) - 1; i++) {
    p = root[id];
    for (j = i; j <= int(d[id].length()) - 1; j++) {
      v = 0;
      we = (*p).head;
      while (we != 0) {
        if (w[we] == d[id][j]) {
          v = a[we];
          break;
        }
        we = trc[we];
      }
      if (v == 0) {
        q = new tnode;
        (*q).head = 0;
        (*q).value = 0;
        edge++;
        a[edge] = q;
        trc[edge] = (*p).head;
        (*p).head = edge;
        w[edge] = d[id][j];
        v = q;
      }
      p = v;
      (*p).value++;
    }
  }
}
void duyet(pnode p, pnode q) {
  int top, day, gt, we, we2;
  top = 1;
  day = 1;
  st[1][0] = p;
  st[1][1] = q;
  while (day <= top) {
    p = st[day][0];
    q = st[day][1];
    if (p != root[0]) {
      if (q == 0)
        gt = 0;
      else
        gt = (*q).value;
      if ((gt < trai) || (gt > phai)) (*p).value = -1;
    }
    we = (*p).head;
    while (we != 0) {
      top++;
      st[top][0] = a[we];
      st[top][1] = 0;
      if (q == 0)
        st[top][1] = 0;
      else {
        we2 = (*q).head;
        while (we2 != 0) {
          if (w[we2] == w[we]) {
            st[top][1] = a[we2];
            break;
          }
          we2 = trc[we2];
        }
      }
      we = trc[we];
    }
    day++;
  }
}
void get(pnode p) {
  if (p != root[0]) {
    if ((*p).value >= 0) ans++;
  }
  int we;
  we = (*p).head;
  while (we != 0) {
    get(a[we]);
    we = trc[we];
  }
}
void xoa(pnode p) {
  int we;
  we = (*p).head;
  while (we != 0) {
    xoa(a[we]);
    we = trc[we];
  }
  delete p;
}
void solve() {
  edge = 0;
  build(0);
  int temp;
  temp = edge;
  int i;
  for (i = 1; i <= n; i++) {
    trai = l[i];
    phai = r[i];
    edge = temp;
    build(i);
    duyet(root[0], root[i]);
    xoa(root[i]);
  }
  ans = 0;
  get(root[0]);
  cout << ans;
}
int main() {
  int i;
  char ch;
  getline(cin, d[0]);
  scanf("%d\n", &n);
  for (i = 1; i <= n; i++) {
    d[i] = "";
    while (true) {
      scanf("%c", &ch);
      if (ch == ' ') break;
      d[i] += ch;
    }
    scanf("%d%d\n", &l[i], &r[i]);
  }
  solve();
  return 0;
}
