#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 222;
const int M = 5e5 + 100;
const int LEN = 105;
const int hashMod = 1e6 + 5;
const int inf = 0x55555555;
const double eps = 1e-8;
const long double leps = 1e-10;
const long long linf = 0x5555555555555555ll;
const double finf = 1e50;
const double pi = acos(-1.0);
const int MOD = 1000000000;
template <class T>
inline T sqr(T x) {
  return x * x;
}
map<string, int> cnt[15];
char buf[N], tmp[N];
int l[15], r[15];
bool check(char *str, int n) {
  for (int i = 0; i < n; i++) {
    if (cnt[i].find(str) == cnt[i].end()) {
      if (l[i] > 0) return false;
    } else {
      int t = cnt[i][str];
      if (t > r[i] || t < l[i]) return false;
    }
  }
  return true;
}
set<string> has;
int main() {
  while (~scanf("%s", buf)) {
    int n;
    scanf("%d", &n);
    has.clear();
    for (int t = 0; t < n; t++) {
      scanf("%s%d%d", tmp, &l[t], &r[t]);
      cnt[t].clear();
      int len = strlen(tmp);
      for (int i = 0; i < len; i++) {
        for (int j = i + 1; j <= len; j++) {
          char tc = tmp[j];
          tmp[j] = 0;
          if (cnt[t].find(tmp + i) == cnt[t].end()) cnt[t][tmp + i] = 0;
          cnt[t][tmp + i]++;
          tmp[j] = tc;
        }
      }
    }
    int len = strlen(buf);
    for (int i = 0; i < len; i++) {
      for (int j = i + 1; j <= len; j++) {
        char tc = buf[j];
        buf[j] = 0;
        if (check(buf + i, n)) {
          has.insert(buf + i);
        }
        buf[j] = tc;
      }
    }
    printf("%d\n", ((int)has.size()));
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 222;
const int M = 5e5 + 100;
const int LEN = 105;
const int hashMod = 1e6 + 5;
const int inf = 0x55555555;
const double eps = 1e-8;
const long double leps = 1e-10;
const long long linf = 0x5555555555555555ll;
const double finf = 1e50;
const double pi = acos(-1.0);
const int MOD = 1000000000;
template <class T>
inline T sqr(T x) {
  return x * x;
}
map<string, int> cnt[15];
char buf[N], tmp[N];
int l[15], r[15];
bool check(char *str, int n) {
  for (int i = 0; i < n; i++) {
    if (cnt[i].find(str) == cnt[i].end()) {
      if (l[i] > 0) return false;
    } else {
      int t = cnt[i][str];
      if (t > r[i] || t < l[i]) return false;
    }
  }
  return true;
}
set<string> has;
int main() {
  while (~scanf("%s", buf)) {
    int n;
    scanf("%d", &n);
    has.clear();
    for (int t = 0; t < n; t++) {
      scanf("%s%d%d", tmp, &l[t], &r[t]);
      cnt[t].clear();
      int len = strlen(tmp);
      for (int i = 0; i < len; i++) {
        for (int j = i + 1; j <= len; j++) {
          char tc = tmp[j];
          tmp[j] = 0;
          if (cnt[t].find(tmp + i) == cnt[t].end()) cnt[t][tmp + i] = 0;
          cnt[t][tmp + i]++;
          tmp[j] = tc;
        }
      }
    }
    int len = strlen(buf);
    for (int i = 0; i < len; i++) {
      for (int j = i + 1; j <= len; j++) {
        char tc = buf[j];
        buf[j] = 0;
        if (check(buf + i, n)) {
          has.insert(buf + i);
        }
        buf[j] = tc;
      }
    }
    printf("%d\n", ((int)has.size()));
  }
  return 0;
}

```