#include <bits/stdc++.h>
using namespace std;
int n, m;
char str[1000010];
map<string, bool> M;
string sub[100010];
int num = 0;
bool vis[100010];
char str1[100010], str2[100010];
int P[100010];
bool MEET(int which, string p, int l, int r) {
  int cnt = 0;
  if (sub[which].size() <= p.size()) {
    for (int i = 0; i < p.size(); i++) str1[i + 1] = p[i];
    for (int i = 0; i < sub[which].size(); i++) str2[i + 1] = sub[which][i];
    memset(P, 0, sizeof(P));
    P[1] = 0;
    int l1 = p.size();
    int l2 = sub[which].size();
    int j = 0;
    for (int i = 2; i <= l2; i++) {
      while (j > 0 && j + 1 <= l2 && str2[j + 1] != str2[i]) j = P[j];
      if (j + 1 <= l2 && str2[j + 1] == str2[i]) j = j + 1;
      P[i] = j;
    }
    j = 0;
    for (int i = 1; i <= l1; i++) {
      while (j > 0 && j + 1 <= l2 && str2[j + 1] != str1[i]) j = P[j];
      if (j + 1 <= l2 && str2[j + 1] == str1[i]) j = j + 1;
      if (j == l2) {
        j = P[j];
        cnt++;
      }
    }
  }
  return (cnt <= r && cnt >= l);
}
int main() {
  int Cas = 0;
  while (scanf("%s", str) != EOF) {
    memset(vis, true, sizeof(vis));
    M.clear();
    int len = strlen(str);
    num = 0;
    for (int i = 0; i < len; i++) {
      string temp = "";
      for (int l = 1; l <= len; l++) {
        if (i + l - 1 >= len) break;
        int ed = i + l - 1;
        temp += str[ed];
        if (!M[temp]) {
          M[temp] = 1;
          num++;
          sub[num] = temp;
        }
      }
    }
    scanf("%d", &n);
    string p;
    int l, r;
    int ans = num;
    for (int i = 1; i <= n; i++) {
      cin >> p >> l >> r;
      for (int j = 1; j <= num; j++)
        if (vis[j])
          if (!MEET(j, p, l, r)) {
            vis[j] = false;
            ans--;
          }
    }
    cout << ans << endl;
  }
}

### Prompt

Please provide a cpp coded solution 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;
int n, m;
char str[1000010];
map<string, bool> M;
string sub[100010];
int num = 0;
bool vis[100010];
char str1[100010], str2[100010];
int P[100010];
bool MEET(int which, string p, int l, int r) {
  int cnt = 0;
  if (sub[which].size() <= p.size()) {
    for (int i = 0; i < p.size(); i++) str1[i + 1] = p[i];
    for (int i = 0; i < sub[which].size(); i++) str2[i + 1] = sub[which][i];
    memset(P, 0, sizeof(P));
    P[1] = 0;
    int l1 = p.size();
    int l2 = sub[which].size();
    int j = 0;
    for (int i = 2; i <= l2; i++) {
      while (j > 0 && j + 1 <= l2 && str2[j + 1] != str2[i]) j = P[j];
      if (j + 1 <= l2 && str2[j + 1] == str2[i]) j = j + 1;
      P[i] = j;
    }
    j = 0;
    for (int i = 1; i <= l1; i++) {
      while (j > 0 && j + 1 <= l2 && str2[j + 1] != str1[i]) j = P[j];
      if (j + 1 <= l2 && str2[j + 1] == str1[i]) j = j + 1;
      if (j == l2) {
        j = P[j];
        cnt++;
      }
    }
  }
  return (cnt <= r && cnt >= l);
}
int main() {
  int Cas = 0;
  while (scanf("%s", str) != EOF) {
    memset(vis, true, sizeof(vis));
    M.clear();
    int len = strlen(str);
    num = 0;
    for (int i = 0; i < len; i++) {
      string temp = "";
      for (int l = 1; l <= len; l++) {
        if (i + l - 1 >= len) break;
        int ed = i + l - 1;
        temp += str[ed];
        if (!M[temp]) {
          M[temp] = 1;
          num++;
          sub[num] = temp;
        }
      }
    }
    scanf("%d", &n);
    string p;
    int l, r;
    int ans = num;
    for (int i = 1; i <= n; i++) {
      cin >> p >> l >> r;
      for (int j = 1; j <= num; j++)
        if (vis[j])
          if (!MEET(j, p, l, r)) {
            vis[j] = false;
            ans--;
          }
    }
    cout << ans << endl;
  }
}

```