#include <bits/stdc++.h>
using namespace std;
const int MAX = 2001;
const long long B = 139;
char s[MAX];
char group[10][MAX];
int con[10][2];
int N;
vector<long long> cnt[10];
int main() {
  int i, j;
  scanf("%s%d", s, &N);
  for (i = 0; i < N; ++i) {
    scanf("%s%d%d", group[i], &con[i][0], &con[i][1]);
    cnt[i].reserve(strlen(group[i]) * strlen(group[i]));
  }
  for (int t = 0; t < N; ++t) {
    int len = strlen(group[t]);
    for (i = 0; i < len; ++i) {
      long long hash = 0;
      for (j = i; j < len; ++j) {
        hash = hash * B + group[t][j];
        cnt[t].push_back(hash);
      }
    }
    sort(cnt[t].begin(), cnt[t].end());
  }
  int ans = 0, len = strlen(s);
  set<long long> mark;
  for (i = 0; i < len; ++i) {
    long long hash = 0;
    for (j = i; j < len; ++j) {
      hash = hash * B + s[j];
      bool ok = true;
      for (int t = 0; t < N && ok; ++t) {
        int c = upper_bound(cnt[t].begin(), cnt[t].end(), hash) -
                lower_bound(cnt[t].begin(), cnt[t].end(), hash);
        ok = c >= con[t][0] && c <= con[t][1];
      }
      if (ok && mark.insert(hash).second) {
        ++ans;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 MAX = 2001;
const long long B = 139;
char s[MAX];
char group[10][MAX];
int con[10][2];
int N;
vector<long long> cnt[10];
int main() {
  int i, j;
  scanf("%s%d", s, &N);
  for (i = 0; i < N; ++i) {
    scanf("%s%d%d", group[i], &con[i][0], &con[i][1]);
    cnt[i].reserve(strlen(group[i]) * strlen(group[i]));
  }
  for (int t = 0; t < N; ++t) {
    int len = strlen(group[t]);
    for (i = 0; i < len; ++i) {
      long long hash = 0;
      for (j = i; j < len; ++j) {
        hash = hash * B + group[t][j];
        cnt[t].push_back(hash);
      }
    }
    sort(cnt[t].begin(), cnt[t].end());
  }
  int ans = 0, len = strlen(s);
  set<long long> mark;
  for (i = 0; i < len; ++i) {
    long long hash = 0;
    for (j = i; j < len; ++j) {
      hash = hash * B + s[j];
      bool ok = true;
      for (int t = 0; t < N && ok; ++t) {
        int c = upper_bound(cnt[t].begin(), cnt[t].end(), hash) -
                lower_bound(cnt[t].begin(), cnt[t].end(), hash);
        ok = c >= con[t][0] && c <= con[t][1];
      }
      if (ok && mark.insert(hash).second) {
        ++ans;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```