#include <bits/stdc++.h>
using namespace std;
char PP[2500];
char TT[2500][2500];
char GG[2500];
int l[2500], r[2500];
map<string, int> Hash;
int N;
int m, n;
char P[2500], T[2500];
int pi[2500];
int MakePI() {
  int i, k;
  k = pi[1] = 0;
  for (i = 2; i <= m; i++) {
    while (k > 0 && P[i] != P[k + 1]) k = pi[k];
    if (P[i] == P[k + 1]) k++;
    pi[i] = k;
  }
  return 0;
}
int Find() {
  int i, k, cnt;
  k = cnt = 0;
  for (i = 1; i <= n; i++) {
    while (k > 0 && T[i] != P[k + 1]) k = pi[k];
    if (T[i] == P[k + 1]) k++;
    if (k == m) {
      cnt++;
      k = pi[k];
    }
  }
  return cnt;
}
int find(char t[], char p[]) {
  m = strlen(p);
  n = strlen(t);
  strcpy(P + 1, p);
  strcpy(T + 1, t);
  MakePI();
  return Find();
}
int main() {
  scanf("%s", PP);
  scanf("%d", &N);
  int i, k, j;
  for (i = 0; i < N; i++) {
    scanf("%s%d%d", TT[i], &l[i], &r[i]);
  }
  int rlt = 0;
  int m;
  for (i = 0; PP[i]; i++) {
    string st = "";
    for (k = i, m = 0; PP[k]; k++, m++) {
      st += PP[k];
      GG[m] = PP[k];
      GG[m + 1] = 0;
      for (j = 0; j < N; j++) {
        int t = find(TT[j], GG);
        if (t > r[j] || t < l[j]) break;
      }
      if (j < N) continue;
      if (!Hash.count(st)) {
        Hash[st] = 1;
        rlt++;
      }
    }
  }
  cout << rlt << endl;
}

### 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;
char PP[2500];
char TT[2500][2500];
char GG[2500];
int l[2500], r[2500];
map<string, int> Hash;
int N;
int m, n;
char P[2500], T[2500];
int pi[2500];
int MakePI() {
  int i, k;
  k = pi[1] = 0;
  for (i = 2; i <= m; i++) {
    while (k > 0 && P[i] != P[k + 1]) k = pi[k];
    if (P[i] == P[k + 1]) k++;
    pi[i] = k;
  }
  return 0;
}
int Find() {
  int i, k, cnt;
  k = cnt = 0;
  for (i = 1; i <= n; i++) {
    while (k > 0 && T[i] != P[k + 1]) k = pi[k];
    if (T[i] == P[k + 1]) k++;
    if (k == m) {
      cnt++;
      k = pi[k];
    }
  }
  return cnt;
}
int find(char t[], char p[]) {
  m = strlen(p);
  n = strlen(t);
  strcpy(P + 1, p);
  strcpy(T + 1, t);
  MakePI();
  return Find();
}
int main() {
  scanf("%s", PP);
  scanf("%d", &N);
  int i, k, j;
  for (i = 0; i < N; i++) {
    scanf("%s%d%d", TT[i], &l[i], &r[i]);
  }
  int rlt = 0;
  int m;
  for (i = 0; PP[i]; i++) {
    string st = "";
    for (k = i, m = 0; PP[k]; k++, m++) {
      st += PP[k];
      GG[m] = PP[k];
      GG[m + 1] = 0;
      for (j = 0; j < N; j++) {
        int t = find(TT[j], GG);
        if (t > r[j] || t < l[j]) break;
      }
      if (j < N) continue;
      if (!Hash.count(st)) {
        Hash[st] = 1;
        rlt++;
      }
    }
  }
  cout << rlt << endl;
}

```