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;
template <class T>
int getbit(int i, T X) {
  return (X & (1 << (i - 1)));
}
template <class T>
T onbit(int i, T X) {
  return (X | (1 << (i - 1)));
}
template <class T>
T offbit(int i, T X) {
  return (X | (1 << (i - 1)) - (1 << (i - 1)));
}
template <class T>
T sqr(T x) {
  return (x * x);
}
template <class T>
T cube(T x) {
  return (x * x * x);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int csx[4] = {0, 0, -1, 1};
int csy[4] = {-1, 1, 0, 0};
const int MOD = 1000000007;
const int inf = 1e9;
const long long linf = 1e18;
const int maxn = 15;
struct node {
  int cnt, f, id;
  char v;
  node *c[26];
};
int n, C = 0;
string t, p[maxn + 5];
int l[maxn + 5], r[maxn + 5];
node *root;
void enter() {
  cin >> t;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> l[i] >> r[i];
  }
}
void add(node *&p, char v) {
  p = new node();
  p->cnt = 0;
  p->f = 0;
  p->v = v;
  for (int i = 0; i <= 25; i++) p->c[i] = NULL;
}
void insert_t(string s) {
  node *cur = root;
  int sz = s.size();
  for (int i = 0; i < sz; i++) {
    if (cur->c[s[i] - 'a'] == NULL) {
      add(cur->c[s[i] - 'a'], s[i]);
    }
    cur = cur->c[s[i] - 'a'];
  }
}
void search_t(string s) {
  int sz = s.size();
  node *cur = root;
  for (int i = 0; i < sz; i++) {
    if (cur->c[s[i] - 'a'] == NULL) return;
    cur->c[s[i] - 'a']->cnt++;
    cur = cur->c[s[i] - 'a'];
  }
}
void reset(node *&p) {
  p->cnt = 0;
  for (int i = 0; i <= 25; i++)
    if (p->c[i] != NULL) reset(p->c[i]);
}
int countResult(node *&p) {
  int ans = 0;
  for (int i = 0; i <= 25; i++)
    if (p->c[i] != NULL) {
      if (p->c[i]->f == n)
        ans += 1 + countResult(p->c[i]);
      else
        ans += countResult(p->c[i]);
    }
  return ans;
}
string toString(int d) {
  string ans = "";
  do {
    ans = (char)(d % 10 + '0') + ans;
    d /= 10;
  } while (d > 0);
  return ans;
}
string ff(int id, char kt) {
  string ans = "";
  ans = "(" + toString(id) + "," + kt + ")";
  return ans;
}
void updateF(node *&p, int id) {
  for (int i = 0; i <= 25; i++)
    if (p->c[i] != NULL) {
      if ((p->c[i]->cnt >= l[id]) && (p->c[i]->cnt <= r[id])) p->c[i]->f++;
      updateF(p->c[i], id);
    }
}
void dfs(node *&p) {
  C++;
  p->id = C;
  for (int i = 0; i <= 25; i++)
    if (p->c[i] != NULL) {
      dfs(p->c[i]);
      cout << ff(p->c[i]->id, p->c[i]->v) << "->" << ff(p->id, p->v);
      cout << endl;
    }
}
void solve() {
  add(root, '?');
  int sz = t.size();
  for (int i = 0; i < sz; i++) {
    insert_t(t.substr(i, sz - i));
  }
  for (int i = 1; i <= n; i++) {
    sz = p[i].size();
    reset(root);
    for (int j = 0; j < sz; j++) {
      search_t(p[i].substr(j, sz - j));
    }
    updateF(root, i);
  }
  int res = countResult(root);
  cout << res << endl;
}
int main() {
  enter();
  solve();
  return 0;
}
