#include <bits/stdc++.h>
using namespace std;
const double EPS = 1E-9;
const long long INF = 1000000000;
const long long INF64 = (long long)1E18;
const double PI = 3.1415926535897932384626433832795;
typedef struct {
  int a;
  double p;
} tpo;
int *pref(int *p, char *s, int len) {
  p[0] = -1;
  int m = len, k = -1;
  for (int j = 1; j < m; ++j) {
    while ((k >= 0) && (s[k + 1] != s[j])) k = p[k];
    if (s[k + 1] == s[j]) ++k;
    p[j] = k;
  }
  return p;
}
int kmp(char *t, int tl, char *s, int *p, int sl) {
  int k = -1;
  if (tl < 1) return 0;
  if (tl < sl) return 0;
  int sum = 0;
  for (int j = 0; j < tl; ++j) {
    while ((k >= 0) && (s[k + 1] != t[j])) k = p[k];
    if (s[k + 1] == t[j]) ++k;
    if (k == sl - 1) {
      sum++;
      k = p[k];
    }
  }
  return sum;
}
typedef struct {
  char s[300];
  int l, r, len;
} tp;
int main() {
  long long i, j, n, m;
  char s[300];
  tp *t = new tp[12];
  cin >> s >> n;
  for (i = 0; i < n; ++i) {
    memset(&t[i].s, 0, 300);
    cin >> t[i].s >> t[i].l >> t[i].r;
    t[i].len = strlen(t[i].s);
  }
  int *p = new int[2000];
  int len, slen = strlen(s), res = 0;
  for (len = 1; len <= slen; ++len) {
    for (i = 0; i + len <= slen; ++i) {
      memset(p, 0, sizeof(int) * (len + 1));
      pref(p, s + i, len);
      if (kmp(s, i + len - 1, s + i, p, len) <= 0) {
        bool bOk = true;
        for (j = 0; j < n && bOk; ++j) {
          int cur = kmp(t[j].s, t[j].len, s + i, p, len);
          if (!(cur >= t[j].l && cur <= t[j].r)) bOk = false;
        }
        if (bOk) res++;
      }
    }
  }
  cout << res;
  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 double EPS = 1E-9;
const long long INF = 1000000000;
const long long INF64 = (long long)1E18;
const double PI = 3.1415926535897932384626433832795;
typedef struct {
  int a;
  double p;
} tpo;
int *pref(int *p, char *s, int len) {
  p[0] = -1;
  int m = len, k = -1;
  for (int j = 1; j < m; ++j) {
    while ((k >= 0) && (s[k + 1] != s[j])) k = p[k];
    if (s[k + 1] == s[j]) ++k;
    p[j] = k;
  }
  return p;
}
int kmp(char *t, int tl, char *s, int *p, int sl) {
  int k = -1;
  if (tl < 1) return 0;
  if (tl < sl) return 0;
  int sum = 0;
  for (int j = 0; j < tl; ++j) {
    while ((k >= 0) && (s[k + 1] != t[j])) k = p[k];
    if (s[k + 1] == t[j]) ++k;
    if (k == sl - 1) {
      sum++;
      k = p[k];
    }
  }
  return sum;
}
typedef struct {
  char s[300];
  int l, r, len;
} tp;
int main() {
  long long i, j, n, m;
  char s[300];
  tp *t = new tp[12];
  cin >> s >> n;
  for (i = 0; i < n; ++i) {
    memset(&t[i].s, 0, 300);
    cin >> t[i].s >> t[i].l >> t[i].r;
    t[i].len = strlen(t[i].s);
  }
  int *p = new int[2000];
  int len, slen = strlen(s), res = 0;
  for (len = 1; len <= slen; ++len) {
    for (i = 0; i + len <= slen; ++i) {
      memset(p, 0, sizeof(int) * (len + 1));
      pref(p, s + i, len);
      if (kmp(s, i + len - 1, s + i, p, len) <= 0) {
        bool bOk = true;
        for (j = 0; j < n && bOk; ++j) {
          int cur = kmp(t[j].s, t[j].len, s + i, p, len);
          if (!(cur >= t[j].l && cur <= t[j].r)) bOk = false;
        }
        if (bOk) res++;
      }
    }
  }
  cout << res;
  return 0;
}

```