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 maxn = 2e3 + 10, maxm = 1e3 + 10;
const long long mod = 1e9 + 12341;
int n, m;
const unsigned long long decm = 131;
unsigned long long pw[maxn];
void init(int n = maxn - 5) {
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * decm;
}
class strhash {
 public:
  unsigned long long hs[maxn], len;
  void calhs(char *s, int n) {
    len = n;
    for (int i = 0; i <= len - 1; ++i) hs[i + 1] = hs[i] * decm + s[i];
  }
  inline unsigned long long geths(int a, int b) {
    return hs[b] - hs[a - 1] * pw[b - a + 1];
  }
} mps;
unordered_map<unsigned long long, int> cnt;
unordered_set<unsigned long long> vis, s1, s2;
char s[maxn];
int l, r;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init(maxn - 5);
  cin >> s >> n;
  m = strlen(s);
  mps.calhs(s, m);
  for (int i = 1; i <= m; ++i)
    for (int j = i; j <= m; ++j) {
      unsigned long long x = mps.geths(i, j);
      if (!vis.count(x)) {
        vis.insert(x);
        s1.insert(x);
      }
    }
  if (n == 0) {
    cout << vis.size();
    return 0;
  }
  int ans = 0;
  unordered_set<unsigned long long> &sub = s1, &tmp = s2;
  for (int i = 1; i <= n; ++i) {
    cin >> s >> l >> r;
    int len = strlen(s);
    cnt.clear();
    mps.calhs(s, len);
    for (int j = 1; j <= len; ++j)
      for (int k = j; k <= len; ++k) {
        if (k - j + 1 > m) break;
        unsigned long long x = mps.geths(j, k);
        if (!vis.count(x)) break;
        cnt[x]++;
      }
    if (l == 0) {
      for (auto x : cnt)
        if (sub.count(x.first) && x.second > r) sub.erase(x.first);
      if (i == n) ans = sub.size();
      continue;
    }
    for (auto x : cnt)
      if (sub.count(x.first) && x.second >= l && x.second <= r)
        tmp.insert(x.first);
    sub.clear();
    if (i == n) ans = tmp.size();
    swap(tmp, sub);
  }
  cout << ans << '\n';
}
