#include <bits/stdc++.h>
using namespace std;
char m[12][2100];
int h1[12][2100], h2[12][2100], p1[2100], p2[2100], a[12], b[12];
set<long long> s;
long long hs[12][10007 + 1000];
int hv[12][10007 + 1000], vs, v[12][10007 + 1000], len[12];
long long gethash(int k, int i, int j) {
  int t1 = ((long long)(h1[k][j] - h1[k][i]) * p1[2100 - 1 - j] % 1000000009 +
            1000000009) %
           1000000009;
  int t2 = ((long long)(h2[k][j] - h2[k][i]) * p2[2100 - 1 - j] % 726731231 +
            726731231) %
           726731231;
  return (long long)t1 * 7326731231 + t2;
}
void add(int k, long long t) {
  int i;
  for (i = t % 10007; v[k][i] == vs && hs[k][i] != t; i++)
    ;
  if (v[k][i] == vs)
    hv[k][i]++;
  else {
    v[k][i] = vs;
    hs[k][i] = t;
    hv[k][i] = 1;
  }
}
int get(int k, long long t) {
  int i;
  for (i = t % 10007; v[k][i] == vs && hs[k][i] != t; i++)
    ;
  if (v[k][i] == vs)
    return hv[k][i];
  else
    return 0;
}
int main() {
  int i, j, l, r, n, t;
  long long k;
  for (p1[0] = 1, i = 1; i < 2100;
       p1[i] = (long long)p1[i - 1] * 123121 % 1000000009, i++)
    ;
  for (p2[0] = 1, i = 1; i < 2100;
       p2[i] = (long long)p2[i - 1] * 717713 % 726731231, i++)
    ;
  for (scanf("%s%d", &m[0], &n), i = 1; i <= n;
       scanf("%s%d%d", m[i], &a[i], &b[i]), i++)
    ;
  for (i = 0; i <= n; len[i] = strlen(m[i]), i++) {
    for (j = 0; m[i][j];
         h1[i][j + 1] = (h1[i][j] + (long long)p1[j] * m[i][j]) % 1000000009,
        j++)
      ;
    for (j = 0; m[i][j];
         h2[i][j + 1] = (h2[i][j] + (long long)p2[j] * m[i][j]) % 726731231,
        j++)
      ;
  }
  vs = 1;
  for (r = 0, l = 1; l <= len[0]; vs++, l++) {
    s.clear();
    for (i = 1; i <= n; i++)
      for (j = 0; j + l <= len[i]; add(i, gethash(i, j, j + l)), j++)
        ;
    for (j = 0; j + l <= len[0]; j++) {
      k = gethash(0, j, j + l);
      if (s.find(k) != s.end()) continue;
      for (i = 1; i <= n; i++) {
        t = get(i, k);
        if (t < a[i] || t > b[i]) break;
      }
      if (i > n) s.insert(k);
    }
    r += s.size();
  }
  printf("%d\n", r);
  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;
char m[12][2100];
int h1[12][2100], h2[12][2100], p1[2100], p2[2100], a[12], b[12];
set<long long> s;
long long hs[12][10007 + 1000];
int hv[12][10007 + 1000], vs, v[12][10007 + 1000], len[12];
long long gethash(int k, int i, int j) {
  int t1 = ((long long)(h1[k][j] - h1[k][i]) * p1[2100 - 1 - j] % 1000000009 +
            1000000009) %
           1000000009;
  int t2 = ((long long)(h2[k][j] - h2[k][i]) * p2[2100 - 1 - j] % 726731231 +
            726731231) %
           726731231;
  return (long long)t1 * 7326731231 + t2;
}
void add(int k, long long t) {
  int i;
  for (i = t % 10007; v[k][i] == vs && hs[k][i] != t; i++)
    ;
  if (v[k][i] == vs)
    hv[k][i]++;
  else {
    v[k][i] = vs;
    hs[k][i] = t;
    hv[k][i] = 1;
  }
}
int get(int k, long long t) {
  int i;
  for (i = t % 10007; v[k][i] == vs && hs[k][i] != t; i++)
    ;
  if (v[k][i] == vs)
    return hv[k][i];
  else
    return 0;
}
int main() {
  int i, j, l, r, n, t;
  long long k;
  for (p1[0] = 1, i = 1; i < 2100;
       p1[i] = (long long)p1[i - 1] * 123121 % 1000000009, i++)
    ;
  for (p2[0] = 1, i = 1; i < 2100;
       p2[i] = (long long)p2[i - 1] * 717713 % 726731231, i++)
    ;
  for (scanf("%s%d", &m[0], &n), i = 1; i <= n;
       scanf("%s%d%d", m[i], &a[i], &b[i]), i++)
    ;
  for (i = 0; i <= n; len[i] = strlen(m[i]), i++) {
    for (j = 0; m[i][j];
         h1[i][j + 1] = (h1[i][j] + (long long)p1[j] * m[i][j]) % 1000000009,
        j++)
      ;
    for (j = 0; m[i][j];
         h2[i][j + 1] = (h2[i][j] + (long long)p2[j] * m[i][j]) % 726731231,
        j++)
      ;
  }
  vs = 1;
  for (r = 0, l = 1; l <= len[0]; vs++, l++) {
    s.clear();
    for (i = 1; i <= n; i++)
      for (j = 0; j + l <= len[i]; add(i, gethash(i, j, j + l)), j++)
        ;
    for (j = 0; j + l <= len[0]; j++) {
      k = gethash(0, j, j + l);
      if (s.find(k) != s.end()) continue;
      for (i = 1; i <= n; i++) {
        t = get(i, k);
        if (t < a[i] || t > b[i]) break;
      }
      if (i > n) s.insert(k);
    }
    r += s.size();
  }
  printf("%d\n", r);
  return 0;
}

```