#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 13, N = 11, L = 205;
int h[N][L], l[N], r[N], p[L];
string a[N];
int get(string s) {
  int ans = 0;
  for (int i = 0; i < ((int)(s).size()); i++)
    ans = (ans * 27ll + s[i] - 'a' + 1) % mod;
  return ans;
}
int gethash(int i, int x, int y) {
  if (x == 0) return h[i][y];
  return (h[i][y] + mod - 1ll * h[i][x - 1] * p[y - x + 1] % mod) % mod;
}
int main() {
  p[0] = 1;
  for (int i = 1; i < L; i++) p[i] = p[i - 1] * 27ll % mod;
  string s;
  cin >> s;
  vector<pair<int, int> > v;
  for (int i = 0; i < ((int)(s).size()); i++)
    for (int j = i; j < ((int)(s).size()); j++)
      v.push_back(pair<int, int>(get(s.substr(i, j - i + 1)), j - i + 1));
  sort(v.begin(), v.end());
  vector<pair<int, int> > t;
  t.push_back(v[0]);
  for (int i = 1; i < ((int)(v).size()); i++)
    if (v[i] != v[i - 1]) t.push_back(v[i]);
  v = t;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> l[i] >> r[i];
    h[i][0] = a[i][0] - 'a' + 1;
    for (int j = 1; j < ((int)(a[i]).size()); j++)
      h[i][j] = (h[i][j - 1] * 27ll + a[i][j] - 'a' + 1) % mod;
  }
  int ans = 0;
  for (int i = 0; i < ((int)(v).size()); i++) {
    bool ok = true;
    for (int j = 0; j < n; j++) {
      int cnt = 0;
      for (int k = 0; k + v[i].second <= ((int)(a[j]).size()); k++)
        if (gethash(j, k, k + v[i].second - 1) == v[i].first) cnt++;
      if (cnt < l[j] || cnt > r[j]) {
        ok = false;
        break;
      }
    }
    if (ok) ans++;
  }
  printf("%d", ans);
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 13, N = 11, L = 205;
int h[N][L], l[N], r[N], p[L];
string a[N];
int get(string s) {
  int ans = 0;
  for (int i = 0; i < ((int)(s).size()); i++)
    ans = (ans * 27ll + s[i] - 'a' + 1) % mod;
  return ans;
}
int gethash(int i, int x, int y) {
  if (x == 0) return h[i][y];
  return (h[i][y] + mod - 1ll * h[i][x - 1] * p[y - x + 1] % mod) % mod;
}
int main() {
  p[0] = 1;
  for (int i = 1; i < L; i++) p[i] = p[i - 1] * 27ll % mod;
  string s;
  cin >> s;
  vector<pair<int, int> > v;
  for (int i = 0; i < ((int)(s).size()); i++)
    for (int j = i; j < ((int)(s).size()); j++)
      v.push_back(pair<int, int>(get(s.substr(i, j - i + 1)), j - i + 1));
  sort(v.begin(), v.end());
  vector<pair<int, int> > t;
  t.push_back(v[0]);
  for (int i = 1; i < ((int)(v).size()); i++)
    if (v[i] != v[i - 1]) t.push_back(v[i]);
  v = t;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> l[i] >> r[i];
    h[i][0] = a[i][0] - 'a' + 1;
    for (int j = 1; j < ((int)(a[i]).size()); j++)
      h[i][j] = (h[i][j - 1] * 27ll + a[i][j] - 'a' + 1) % mod;
  }
  int ans = 0;
  for (int i = 0; i < ((int)(v).size()); i++) {
    bool ok = true;
    for (int j = 0; j < n; j++) {
      int cnt = 0;
      for (int k = 0; k + v[i].second <= ((int)(a[j]).size()); k++)
        if (gethash(j, k, k + v[i].second - 1) == v[i].first) cnt++;
      if (cnt < l[j] || cnt > r[j]) {
        ok = false;
        break;
      }
    }
    if (ok) ans++;
  }
  printf("%d", ans);
}

```