#include <bits/stdc++.h>
using namespace std;
char p[2003];
char s[13][2003];
struct prav {
  int l, r;
};
prav b[13];
struct node {
  int c;
  map<char, int> next;
};
node a[2000003];
int coun;
void dfs(int x) {
  for (int i = 0; i < 26; i++)
    if (a[x].next.find(i) != a[x].next.end()) {
      dfs(a[x].next[i]);
      a[x].next.erase(i);
    }
  a[x].c = 0;
}
void build_bor(int ind) {
  int i, j;
  int x;
  dfs(0);
  coun = 1;
  for (i = 0; s[ind][i]; i++) {
    x = 0;
    for (j = i; s[ind][j]; j++) {
      if (a[x].next.find(s[ind][j] - 'a') == a[x].next.end()) {
        a[x].next.insert(std::pair<char, int>(s[ind][j] - 'a', coun));
        x = coun;
        coun++;
      } else
        x = a[x].next[s[ind][j] - 'a'];
      a[x].c++;
    }
  }
}
bool ok[2000][2000];
int main() {
  scanf("%s", p);
  int n;
  scanf("%d", &n);
  int i, j, k;
  for (i = 0; i < n; i++) {
    scanf("%s", s[i]);
    scanf("%d %d", &b[i].l, &b[i].r);
  }
  int x;
  for (j = 0; p[j]; j++)
    for (k = j; p[k]; k++) ok[j][k] = true;
  for (i = 0; i < n; i++) {
    build_bor(i);
    for (j = 0; p[j]; j++) {
      x = 0;
      for (k = j; p[k]; k++) {
        if (a[x].next.find(p[k] - 'a') == a[x].next.end()) break;
        x = a[x].next[p[k] - 'a'];
        if (a[x].c < b[i].l || a[x].c > b[i].r) ok[j][k] = false;
      }
      while (p[k]) {
        if (b[i].l > 0) ok[j][k] = false;
        k++;
      }
    }
  }
  dfs(0);
  int ans = 0;
  coun = 1;
  for (j = 0; p[j]; j++) {
    x = 0;
    for (k = j; p[k]; k++) {
      if (a[x].next.find(p[k] - 'a') == a[x].next.end()) {
        a[x].next[p[k] - 'a'] = coun;
        x = coun;
        coun++;
      } else
        x = a[x].next[p[k] - 'a'];
      if (ok[j][k] && a[x].c == 0) {
        ans++;
        a[x].c++;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
char p[2003];
char s[13][2003];
struct prav {
  int l, r;
};
prav b[13];
struct node {
  int c;
  map<char, int> next;
};
node a[2000003];
int coun;
void dfs(int x) {
  for (int i = 0; i < 26; i++)
    if (a[x].next.find(i) != a[x].next.end()) {
      dfs(a[x].next[i]);
      a[x].next.erase(i);
    }
  a[x].c = 0;
}
void build_bor(int ind) {
  int i, j;
  int x;
  dfs(0);
  coun = 1;
  for (i = 0; s[ind][i]; i++) {
    x = 0;
    for (j = i; s[ind][j]; j++) {
      if (a[x].next.find(s[ind][j] - 'a') == a[x].next.end()) {
        a[x].next.insert(std::pair<char, int>(s[ind][j] - 'a', coun));
        x = coun;
        coun++;
      } else
        x = a[x].next[s[ind][j] - 'a'];
      a[x].c++;
    }
  }
}
bool ok[2000][2000];
int main() {
  scanf("%s", p);
  int n;
  scanf("%d", &n);
  int i, j, k;
  for (i = 0; i < n; i++) {
    scanf("%s", s[i]);
    scanf("%d %d", &b[i].l, &b[i].r);
  }
  int x;
  for (j = 0; p[j]; j++)
    for (k = j; p[k]; k++) ok[j][k] = true;
  for (i = 0; i < n; i++) {
    build_bor(i);
    for (j = 0; p[j]; j++) {
      x = 0;
      for (k = j; p[k]; k++) {
        if (a[x].next.find(p[k] - 'a') == a[x].next.end()) break;
        x = a[x].next[p[k] - 'a'];
        if (a[x].c < b[i].l || a[x].c > b[i].r) ok[j][k] = false;
      }
      while (p[k]) {
        if (b[i].l > 0) ok[j][k] = false;
        k++;
      }
    }
  }
  dfs(0);
  int ans = 0;
  coun = 1;
  for (j = 0; p[j]; j++) {
    x = 0;
    for (k = j; p[k]; k++) {
      if (a[x].next.find(p[k] - 'a') == a[x].next.end()) {
        a[x].next[p[k] - 'a'] = coun;
        x = coun;
        coun++;
      } else
        x = a[x].next[p[k] - 'a'];
      if (ok[j][k] && a[x].c == 0) {
        ans++;
        a[x].c++;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```