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 n;
string s;
int ls[10], rs[10];
string ps[10];
unordered_map<string, int> cnts[10];
unordered_map<uint64_t, int> hs[10];
bool ok(string s) {
  for (int i = 0; i < n; ++i) {
    auto it = cnts[i].find(s);
    int c = it == cnts[i].end() ? 0 : it->second;
    if (c < ls[i] || c > rs[i]) return 0;
  }
  return 1;
}
void gen(int k) {
  for (int i = 0; i < n; ++i) {
    cnts[i].clear();
    string p = ps[i];
    int x = p.size();
    for (int j = 0; j + k <= x; ++j) ++cnts[i][p.substr(j, k)];
  }
}
const uint64_t Z = 3333331;
void genhs(int k) {
  uint64_t ZZ = 1;
  for (int i = 1; i < k; ++i) ZZ *= Z;
  for (int i = 0; i < n; ++i) {
    hs[i].clear();
    uint64_t cur = 0;
    string p = ps[i];
    int x = p.size();
    if (x < k) continue;
    for (int j = 0; j < k; ++j) cur = Z * cur + p[j];
    ++hs[i][cur];
    for (int j = k; j < x; ++j) {
      cur -= ZZ * p[j - k];
      cur = Z * cur + p[j];
      ++hs[i][cur];
    }
  }
}
bool okh(uint64_t x) {
  for (int i = 0; i < n; ++i) {
    auto it = hs[i].find(x);
    int c = it == hs[i].end() ? 0 : it->second;
    if (c < ls[i] || c > rs[i]) return 0;
  }
  return 1;
}
int main() {
  cin >> s >> n;
  for (int i = 0; i < n; ++i) cin >> ps[i] >> ls[i] >> rs[i];
  for (int i = 0; i < n; ++i) {
    string p = ps[i];
  }
  int x = s.size();
  int rs = 0;
  for (int j = 1; j <= x; ++j) {
    uint64_t ZZ = 1;
    for (int i = 1; i < j; ++i) ZZ *= Z;
    unordered_set<uint64_t> res;
    genhs(j);
    uint64_t cur = 0;
    for (int i = 0; i < j; ++i) cur = Z * cur + s[i];
    if (okh(cur)) res.insert(cur);
    for (int i = j; i < x; ++i) {
      cur -= ZZ * s[i - j];
      cur = Z * cur + s[i];
      if (okh(cur)) res.insert(cur);
    }
    rs += res.size();
  }
  cout << rs << '\n';
}
