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 son[2010 << 1][26], len[2010 << 1], pre[2010 << 1];
int tot, tail;
void init() {
  tail = tot = len[0] = 0;
  pre[0] = -1;
  memset(son, 0, sizeof(son));
}
void add(int x) {
  int p = tail, np = ++tot, q, nq;
  len[np] = len[p] + 1, tail = np;
  while (p != -1 && !son[p][x]) son[p][x] = np, p = pre[p];
  if (p == -1)
    pre[np] = 0;
  else {
    q = son[p][x];
    if (len[p] + 1 == len[q])
      pre[np] = q;
    else {
      nq = ++tot;
      memcpy(son[nq], son[q], sizeof(son[q]));
      len[nq] = len[p] + 1;
      pre[nq] = pre[q];
      pre[q] = pre[np] = nq;
      while (p != -1 && son[p][x] == q) son[p][x] = nq, p = pre[p];
    }
  }
}
char str[2010];
int num[2010 << 1][2010];
int L[11][2010 << 1], R[11][2010 << 1];
int b[2010 << 1];
bool cmp(int x, int y) { return len[x] > len[y]; }
int main() {
  int n;
  int p, le, x, u, v;
  scanf("%s", str);
  init();
  memset(L, -1, sizeof(L));
  memset(R, -1, sizeof(R));
  for (int i = 0; str[i]; ++i) add(str[i] - 'a');
  for (int i = 0; i <= tot; ++i) b[i] = i;
  sort(b, b + tot + 1, cmp);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s%d%d", str, &u, &v);
    memset(num, 0, sizeof(num));
    p = le = 0;
    for (int j = 0; str[j]; ++j) {
      x = str[j] - 'a';
      if (son[p][x]) {
        ++le;
        p = son[p][x];
      } else {
        while (p != -1 && !son[p][x]) p = pre[p];
        if (p != -1) {
          le = len[p] + 1;
          p = son[p][x];
        } else {
          le = p = 0;
        }
      }
      if (p) {
        num[p][le]++;
      }
    }
    for (int j = 0; j < tot; ++j) {
      p = b[j];
      for (int k = len[p]; k > len[pre[p]]; --k) {
        if (k != len[p]) num[p][k] += num[p][k + 1];
        if (num[p][k] >= u && num[p][k] <= v) {
          if (L[i][p] == -1 || L[i][p] > k) L[i][p] = k;
          if (R[i][p] == -1 || L[i][p] < k) R[i][p] = k;
        }
      }
      num[pre[p]][len[pre[p]]] += num[p][len[pre[p]] + 1];
    }
  }
  int ans = 0;
  for (int i = 0; i < tot; ++i) {
    p = b[i];
    v = len[p];
    u = len[pre[p]] + 1;
    int is = 0;
    for (int j = 0; j < n; ++j)
      if (L[j][p] == -1 || R[j][p] == -1) {
        is = 1;
        break;
      }
    if (is) continue;
    for (int j = 0; j < n; ++j) {
      if (u == -1 || u < L[j][p]) u = L[j][p];
      if (v == -1 || v > R[j][p]) v = R[j][p];
    }
    if (u == -1 || v == -1) continue;
    if (v >= u) ans += v - u + 1;
  }
  cout << ans << endl;
}
