#include <bits/stdc++.h>
using namespace std;
const int N = 2020, M = 12, base = 37, MOD = 1000 * 1000 * 1000 + 7;
int sum(int a, int b) {
  a += b;
  if (a < 0)
    a += MOD;
  else if (a >= MOD)
    a -= MOD;
  return a;
}
int mul(int a, int b) { return 1LL * a * b % MOD; }
int n, m, ans, low[N], hi[N];
string s, t[M];
unordered_map<int, bool> done;
unordered_map<int, int> cnt[M];
bool check(int val) {
  if (done[val]) return false;
  done[val] = true;
  for (int i = 0; i < m; i++)
    if (cnt[i][val] < low[i] || hi[i] < cnt[i][val]) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> s >> m;
  n = s.length();
  for (int i = 0; i < m; i++) {
    cin >> t[i] >> low[i] >> hi[i];
    for (int l = 0; l < t[i].length(); l++) {
      int val = 0;
      for (int r = l; r < t[i].length(); r++) {
        val = sum(mul(val, base), t[i][r]);
        cnt[i][val]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    int val = 0;
    for (int j = i; j < n; j++) {
      val = sum(mul(val, base), s[j]);
      ans += check(val);
    }
  }
  cout << ans;
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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 N = 2020, M = 12, base = 37, MOD = 1000 * 1000 * 1000 + 7;
int sum(int a, int b) {
  a += b;
  if (a < 0)
    a += MOD;
  else if (a >= MOD)
    a -= MOD;
  return a;
}
int mul(int a, int b) { return 1LL * a * b % MOD; }
int n, m, ans, low[N], hi[N];
string s, t[M];
unordered_map<int, bool> done;
unordered_map<int, int> cnt[M];
bool check(int val) {
  if (done[val]) return false;
  done[val] = true;
  for (int i = 0; i < m; i++)
    if (cnt[i][val] < low[i] || hi[i] < cnt[i][val]) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> s >> m;
  n = s.length();
  for (int i = 0; i < m; i++) {
    cin >> t[i] >> low[i] >> hi[i];
    for (int l = 0; l < t[i].length(); l++) {
      int val = 0;
      for (int r = l; r < t[i].length(); r++) {
        val = sum(mul(val, base), t[i][r]);
        cnt[i][val]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    int val = 0;
    for (int j = i; j < n; j++) {
      val = sum(mul(val, base), s[j]);
      ans += check(val);
    }
  }
  cout << ans;
  return 0;
}

```