#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e2 + 10, pr = 29;
const long long md = 1ll * 1000000000000037;
int n, ted, lft[20], rgt[20];
long long dpt[10][maxn][maxn], dp[maxn][maxn];
string s, t[20];
map<long long, int> mp[20][maxn], mark[maxn];
void make_hash() {
  for (int i = 0; i < n; i++) {
    for (int l = 0; l < t[i].size(); l++) {
      dpt[i][l][l] = 1ll * (t[i][l] - 'a') % md;
      mp[i][1][dpt[i][l][l]]++;
      for (int r = l + 1; r < t[i].size(); r++) {
        dpt[i][l][r] =
            1ll *
            ((1ll * dpt[i][l][r - 1] * pr) % md + (1ll * (t[i][r] - 'a'))) % md;
        mp[i][r - l + 1][dpt[i][l][r]]++;
      }
    }
  }
}
void count_ted(long long val1, int id) {
  if (n == 0) {
    if (mark[id][val1] == 0) ted++;
    mark[id][val1] = 1;
    return;
  }
  bool ch = true;
  for (int i = 0; i < n; i++)
    if (!(lft[i] <= mp[i][id][val1] && mp[i][id][val1] <= rgt[i])) ch = false;
  if (ch) {
    ted++;
    for (int i = 0; i < n; i++) mp[i][id][val1] = -1;
  }
}
void count_hash() {
  for (int i = 0; i < s.size(); i++) {
    dp[i][i] = 1ll * (s[i] - 'a') % md;
    count_ted(dp[i][i], 1);
    for (int j = i + 1; j < s.size(); j++) {
      dp[i][j] = 1ll * ((1ll * (dp[i][j - 1] * pr) % md) + (s[j] - 'a')) % md;
      count_ted(dp[i][j], j - i + 1);
    }
  }
}
int main() {
  cin >> s >> n;
  for (int i = 0; i < n; i++) cin >> t[i] >> lft[i] >> rgt[i];
  make_hash();
  count_hash();
  cout << ted << '\n';
  return 0;
}

### Prompt

Generate a CPP solution to 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 maxn = 2e2 + 10, pr = 29;
const long long md = 1ll * 1000000000000037;
int n, ted, lft[20], rgt[20];
long long dpt[10][maxn][maxn], dp[maxn][maxn];
string s, t[20];
map<long long, int> mp[20][maxn], mark[maxn];
void make_hash() {
  for (int i = 0; i < n; i++) {
    for (int l = 0; l < t[i].size(); l++) {
      dpt[i][l][l] = 1ll * (t[i][l] - 'a') % md;
      mp[i][1][dpt[i][l][l]]++;
      for (int r = l + 1; r < t[i].size(); r++) {
        dpt[i][l][r] =
            1ll *
            ((1ll * dpt[i][l][r - 1] * pr) % md + (1ll * (t[i][r] - 'a'))) % md;
        mp[i][r - l + 1][dpt[i][l][r]]++;
      }
    }
  }
}
void count_ted(long long val1, int id) {
  if (n == 0) {
    if (mark[id][val1] == 0) ted++;
    mark[id][val1] = 1;
    return;
  }
  bool ch = true;
  for (int i = 0; i < n; i++)
    if (!(lft[i] <= mp[i][id][val1] && mp[i][id][val1] <= rgt[i])) ch = false;
  if (ch) {
    ted++;
    for (int i = 0; i < n; i++) mp[i][id][val1] = -1;
  }
}
void count_hash() {
  for (int i = 0; i < s.size(); i++) {
    dp[i][i] = 1ll * (s[i] - 'a') % md;
    count_ted(dp[i][i], 1);
    for (int j = i + 1; j < s.size(); j++) {
      dp[i][j] = 1ll * ((1ll * (dp[i][j - 1] * pr) % md) + (s[j] - 'a')) % md;
      count_ted(dp[i][j], j - i + 1);
    }
  }
}
int main() {
  cin >> s >> n;
  for (int i = 0; i < n; i++) cin >> t[i] >> lft[i] >> rgt[i];
  make_hash();
  count_hash();
  cout << ted << '\n';
  return 0;
}

```