#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10, MOD = 1000 * 1000 * 1000 + 7;
string Candid[maxn];
bool Can[maxn];
set<string> Set;
string s;
inline string Sub(int l, int r) {
  int i;
  string ans = "";
  for (i = l; i <= r; i++) ans += s[i];
  return ans;
}
void KMPpreprocess(string pattern, vector<int> &back) {
  int i, j;
  i = 0;
  j = -1;
  back.push_back(-1);
  while (i < pattern.size()) {
    while (j >= 0 && pattern[i] != pattern[j]) j = back[j];
    i++;
    j++;
    back.push_back(j);
  }
}
int KMPsearch(vector<int> back, string pattern, string way) {
  int i, j, occ;
  occ = i = j = 0;
  while (i < way.size()) {
    while (j >= 0 && way[i] != pattern[j]) j = back[j];
    i++;
    j++;
    if (j == pattern.size()) {
      occ++;
      j = back[j];
    }
  }
  return occ;
}
int main() {
  int n, left[20], right[20], i, l, r;
  string p[20];
  vector<int> ba;
  cin >> s >> n;
  for (l = 0; l < s.size(); l++) {
    for (r = l; r < s.size(); r++) {
      Set.insert(Sub(l, r));
    }
  }
  set<string>::iterator itr;
  int t;
  t = 0;
  for (itr = Set.begin(); itr != Set.end(); itr++) {
    Candid[t++] = *itr;
  }
  for (i = 0; i < t; i++) Can[i] = true;
  for (i = 0; i < n; i++) cin >> p[i] >> left[i] >> right[i];
  int temp, j;
  for (j = 0; j < t; j++) {
    ba.clear();
    KMPpreprocess(Candid[j], ba);
    for (i = 0; i < n; i++) {
      if (Can[j]) {
        temp = KMPsearch(ba, Candid[j], p[i]);
        if (temp < left[i] || temp > right[i]) Can[j] = false;
      }
    }
  }
  int ans = 0;
  for (i = 0; i < t; i++) {
    if (Can[i]) ans++;
  }
  cout << ans << endl;
  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 maxn = 100000 + 10, MOD = 1000 * 1000 * 1000 + 7;
string Candid[maxn];
bool Can[maxn];
set<string> Set;
string s;
inline string Sub(int l, int r) {
  int i;
  string ans = "";
  for (i = l; i <= r; i++) ans += s[i];
  return ans;
}
void KMPpreprocess(string pattern, vector<int> &back) {
  int i, j;
  i = 0;
  j = -1;
  back.push_back(-1);
  while (i < pattern.size()) {
    while (j >= 0 && pattern[i] != pattern[j]) j = back[j];
    i++;
    j++;
    back.push_back(j);
  }
}
int KMPsearch(vector<int> back, string pattern, string way) {
  int i, j, occ;
  occ = i = j = 0;
  while (i < way.size()) {
    while (j >= 0 && way[i] != pattern[j]) j = back[j];
    i++;
    j++;
    if (j == pattern.size()) {
      occ++;
      j = back[j];
    }
  }
  return occ;
}
int main() {
  int n, left[20], right[20], i, l, r;
  string p[20];
  vector<int> ba;
  cin >> s >> n;
  for (l = 0; l < s.size(); l++) {
    for (r = l; r < s.size(); r++) {
      Set.insert(Sub(l, r));
    }
  }
  set<string>::iterator itr;
  int t;
  t = 0;
  for (itr = Set.begin(); itr != Set.end(); itr++) {
    Candid[t++] = *itr;
  }
  for (i = 0; i < t; i++) Can[i] = true;
  for (i = 0; i < n; i++) cin >> p[i] >> left[i] >> right[i];
  int temp, j;
  for (j = 0; j < t; j++) {
    ba.clear();
    KMPpreprocess(Candid[j], ba);
    for (i = 0; i < n; i++) {
      if (Can[j]) {
        temp = KMPsearch(ba, Candid[j], p[i]);
        if (temp < left[i] || temp > right[i]) Can[j] = false;
      }
    }
  }
  int ans = 0;
  for (i = 0; i < t; i++) {
    if (Can[i]) ans++;
  }
  cout << ans << endl;
  return 0;
}

```