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>
#pragma comment(linker, "/stack:256000000")
using namespace std;
const int INF = 1000000000;
const unsigned long long LINF = 4000000000000000000LL;
const double eps = 1e-9;
void prepare() {}
const unsigned long long G = 3137;
const int maxh = (1 << 23) - 1;
const unsigned long long mod1 = INF + 7;
const unsigned long long mod2 = INF + 9;
struct HT {
  unsigned long long h[maxh + 1];
  short cnt[maxh + 1];
  char flag[maxh + 1];
  char T;
  HT() {
    memset((flag), (0), sizeof(flag));
    T = 0;
  }
  void clear() { T++; }
  void add(unsigned long long k) {
    int x = k & maxh;
    while (flag[x] == T && h[x] != k) (++x) &= maxh;
    if (flag[x] != T) {
      flag[x] = T;
      cnt[x] = 1;
      h[x] = k;
    } else
      cnt[x]++;
  }
  int get(unsigned long long k) {
    int x = k & maxh;
    while (flag[x] == T && h[x] != k) (++x) &= maxh;
    if (flag[x] != T) return 0;
    return cnt[x];
  }
} ht;
char buf[100100];
string s;
string p[15];
int l[15], r[15];
int n;
bool bad[2005][2005];
unsigned long long power1[2010], power2[2010], h1[2010], h2[2010];
string genString() {
  string ret;
  for (int i = 0; i < 2000; i++) ret.push_back(rand() % 26 + 'a');
  return ret;
}
void read() {
  scanf("%s", buf);
  s = buf;
  scanf("%d", &n);
  for (int i = (0); i < ((n)); i++) {
    scanf("%s", buf);
    p[i] = buf;
    scanf("%d%d", &l[i], &r[i]);
  }
}
void gen() {
  n = 10;
  s = genString();
  for (int i = (0); i < ((n)); i++) {
    p[i] = genString();
    l[i] = 0;
    r[i] = INF;
  }
}
void buildHash(string &s) {
  h1[0] = h2[0] = s[0];
  for (int i = 1; i < (int)(s).size(); i++) {
    h1[i] = (h1[i - 1] + power1[i] * s[i]);
  }
}
unsigned long long getHash1(int l, int r) {
  unsigned long long ret = h1[r];
  if (l) ret -= h1[l - 1];
  ret = (ret % mod1 + mod1) % mod1;
  ret *= power1[2010 - 1 - l];
  return ret % mod1;
}
unsigned long long getHash2(int l, int r) {
  unsigned long long ret = h2[r];
  if (l) ret -= h2[l - 1];
  ret = (ret % mod2 + mod2) % mod2;
  ret *= power2[2010 - 1 - l];
  return ret % mod2;
}
unsigned long long getHash(int l, int r) {
  unsigned long long ret = h1[r];
  if (l) ret -= h1[l - 1];
  ret *= power1[2010 - 1 - l];
  return ret;
}
bool solve() {
  power1[0] = power2[0] = 1;
  for (int i = 1; i < 2010; i++) {
    power1[i] = power1[i - 1] * G;
    power2[i] = power2[i - 1] * G % mod2;
  }
  for (int k = (0); k < ((n)); k++) {
    ht.clear();
    buildHash(p[k]);
    for (int len = (1); len < ((int)(p[k]).size() + 1); len++) {
      for (int i = 0; i + len <= (int)(p[k]).size(); i++)
        ht.add(getHash(i, i + len - 1));
    }
    buildHash(s);
    for (int len = (1); len < ((int)(s).size() + 1); len++) {
      for (int i = 0; i + len <= (int)(s).size(); i++) {
        if (bad[len][i]) continue;
        int cnt = ht.get(getHash(i, i + len - 1));
        if (cnt < l[k] || cnt > r[k]) bad[len][i] = true;
      }
    }
  }
  int res = 0;
  buildHash(s);
  ht.clear();
  for (int len = (1); len < ((int)(s).size() + 1); len++) {
    for (int i = 0; i + len <= (int)(s).size(); i++) {
      if (!bad[len][i]) {
        unsigned long long h = getHash(i, i + len - 1);
        if (!ht.get(h)) {
          res++;
          ht.add(h);
        }
      }
    }
  }
  printf("%d\n", res);
  return false;
}
int main() {
  prepare();
  read();
  while (solve())
    ;
  return 0;
}
