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 N = 5e4 + 9;
struct Node {
  int next[26];
  int dep, fail;
  void clear() {
    memset(next, 0, sizeof(next));
    dep = fail = 0;
  }
};
int dp[10][N * 11];
int chef[N * 11];
int sef[N * 11];
pair<int, int> limit[10];
struct SA {
  char s[N];
  Node a[N * 11];
  int sz, en;
  void init() {
    fill(a[0].next, a[0].next + 26, 1);
    a[0].dep = -1;
    en = sz = 1;
    a[sz++].clear();
  }
  void build() {
    init();
    for (int i = 0; s[i]; i++) {
      int key = s[i] - 'a';
      int p = sz++;
      a[p].clear();
      a[p].dep = a[en].dep + 1;
      sef[p] = 1;
      while (!a[en].next[key]) a[en].next[key] = p, en = a[en].fail;
      int p1 = a[en].next[key];
      if (a[en].dep + 1 == a[p1].dep)
        a[p].fail = p1;
      else {
        int p2 = sz++;
        a[p2] = a[p1];
        sef[p2] = 0;
        a[p].fail = a[p1].fail = p2;
        a[p2].dep = a[en].dep + 1;
        while (a[en].next[key] == p1) a[en].next[key] = p2, en = a[en].fail;
      }
      en = p;
    }
  }
  vector<int> to[N << 1];
  void dfs(int x, int ar[]) {
    for (int i = 0; i < to[x].size(); i++) {
      int t = to[x][i];
      dfs(t, ar);
      ar[x] += ar[t];
    }
  }
  void make_tree() {
    for (int i = 2; i < sz; i++) to[a[i].fail].push_back(i);
  }
  void run(char s[], int k) {
    int p = 1;
    for (int i = 0; s[i]; i++) {
      int key = s[i] - 'a';
      while (!a[p].next[key]) p = a[p].fail;
      int p1 = a[p].next[key];
      if (a[p].dep + 1 == a[p1].dep)
        p = p1;
      else {
        int p2 = sz++;
        a[p2] = a[p1];
        sef[p2] = 0;
        a[p2].dep = a[p].dep + 1;
        a[p1].fail = p2;
        while (a[p].next[key] == p1) a[p].next[key] = p2, p = a[p].fail;
        p = p2;
      }
      dp[k][p]++;
    }
  }
} A;
int n;
char ts[N];
int main() {
  scanf("%s", A.s);
  A.build();
  cin >> n;
  for (int i = 0; i < n; i++) {
    int k1, k2;
    scanf("%s%d%d", ts, &k1, &k2);
    limit[i] = make_pair(k1, k2);
    A.run(ts, i);
  }
  A.make_tree();
  for (int i = 0; i < n; i++) {
    A.dfs(1, dp[i]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 2; j < A.sz; j++)
      if (limit[i].first <= dp[i][j] && dp[i][j] <= limit[i].second)
        chef[j] |= 1 << i;
  }
  A.dfs(1, sef);
  long long ans = 0;
  for (int i = 2; i < A.sz; i++)
    if (chef[i] + 1 == (1 << n))
      if (sef[i]) {
        ans += (A.a[i].dep - A.a[A.a[i].fail].dep);
      }
  cout << ans << endl;
  return 0;
}
