#include <bits/stdc++.h>
using namespace std;
char s[2000];
char p[10][2000];
int n, l[10], r[10], ls, lp[10];
pair<int, unsigned long long> vh[2000 * (2000 + 1) / 2],
    S1[2000 * (2000 + 1) / 2], S2[2000 * (2000 + 1) / 2];
pair<int, unsigned long long> H, vh2[2000 * (2000 + 1) / 2];
bool ok[2000 * (2000 + 1) / 2];
bool find1(int sz, pair<int, unsigned long long> &p) {
  if (sz == 0) return false;
  int lo = 0, hi = sz - 1, mi;
  while (lo < hi) {
    mi = (lo + hi) >> 1;
    if (S1[mi] < p)
      lo = mi + 1;
    else
      hi = mi;
  }
  return S1[lo] == p;
}
bool find2(int sz, pair<int, unsigned long long> &p) {
  if (sz == 0) return false;
  int lo = 0, hi = sz - 1, mi;
  while (lo < hi) {
    mi = (lo + hi) >> 1;
    if (S2[mi] < p)
      lo = mi + 1;
    else
      hi = mi;
  }
  return S2[lo] == p;
}
int main() {
  scanf("%s", s);
  ls = strlen(s);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s %d %d", p[i], &l[i], &r[i]);
    lp[i] = strlen(p[i]);
  }
  int sz = 0;
  for (int i = 0; i < ls; ++i) {
    unsigned long long h1 = 0;
    long long h2 = 0;
    int len = 0;
    for (int j = i; j < ls; ++j) {
      h1 = h1 * 103 + s[j];
      ++len;
      vh2[sz++] = make_pair(len, h1);
    }
  }
  sort(vh2, vh2 + sz);
  sz = unique(vh2, vh2 + sz) - vh2;
  memset(ok, true, sizeof ok);
  for (int i = 0; i < n; ++i) {
    int m = 0, n1 = 0, n2 = 0;
    for (int j = 0; j < lp[i]; ++j) {
      unsigned long long h1 = 0;
      int len = 0;
      for (int k = j; k < lp[i]; ++k) {
        h1 = h1 * 103 + p[i][k];
        ++len;
        vh[m].first = len;
        vh[m].second = h1;
        ++m;
      }
    }
    sort(vh, vh + m);
    for (int j = 0; j < m;) {
      int e = j;
      while (e < m && vh[e] == vh[j]) ++e;
      if (l[i] == 0 || r[i] == 0) S2[n2++] = vh[j];
      if (r[i]) {
        if (l[i] == 0 && e - j > r[i]) S1[n1++] = vh[j];
        if (l[i] > 0 && e - j >= l[i] && e - j <= r[i]) S1[n1++] = vh[j];
      }
      j = e;
    }
    sort(S1, S1 + n1);
    sort(S2, S2 + n2);
    n1 = unique(S1, S1 + n1) - S1;
    n2 = unique(S2, S2 + n2) - S2;
    for (int j = 0; j < sz; ++j) {
      if (ok[j]) {
        bool ok2 = false;
        if (r[i] == 0) {
          if (!find2(n2, vh2[j])) ok2 = true;
        } else if (l[i] == 0) {
          if (!find2(n2, H))
            ok2 = true;
          else if (!find1(n1, vh2[j]))
            ok2 = true;
        } else {
          if (find1(n1, vh2[j])) ok2 = true;
        }
        ok[j] = ok2;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < sz; ++i)
    if (ok[i]) ++ans;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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 s[2000];
char p[10][2000];
int n, l[10], r[10], ls, lp[10];
pair<int, unsigned long long> vh[2000 * (2000 + 1) / 2],
    S1[2000 * (2000 + 1) / 2], S2[2000 * (2000 + 1) / 2];
pair<int, unsigned long long> H, vh2[2000 * (2000 + 1) / 2];
bool ok[2000 * (2000 + 1) / 2];
bool find1(int sz, pair<int, unsigned long long> &p) {
  if (sz == 0) return false;
  int lo = 0, hi = sz - 1, mi;
  while (lo < hi) {
    mi = (lo + hi) >> 1;
    if (S1[mi] < p)
      lo = mi + 1;
    else
      hi = mi;
  }
  return S1[lo] == p;
}
bool find2(int sz, pair<int, unsigned long long> &p) {
  if (sz == 0) return false;
  int lo = 0, hi = sz - 1, mi;
  while (lo < hi) {
    mi = (lo + hi) >> 1;
    if (S2[mi] < p)
      lo = mi + 1;
    else
      hi = mi;
  }
  return S2[lo] == p;
}
int main() {
  scanf("%s", s);
  ls = strlen(s);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s %d %d", p[i], &l[i], &r[i]);
    lp[i] = strlen(p[i]);
  }
  int sz = 0;
  for (int i = 0; i < ls; ++i) {
    unsigned long long h1 = 0;
    long long h2 = 0;
    int len = 0;
    for (int j = i; j < ls; ++j) {
      h1 = h1 * 103 + s[j];
      ++len;
      vh2[sz++] = make_pair(len, h1);
    }
  }
  sort(vh2, vh2 + sz);
  sz = unique(vh2, vh2 + sz) - vh2;
  memset(ok, true, sizeof ok);
  for (int i = 0; i < n; ++i) {
    int m = 0, n1 = 0, n2 = 0;
    for (int j = 0; j < lp[i]; ++j) {
      unsigned long long h1 = 0;
      int len = 0;
      for (int k = j; k < lp[i]; ++k) {
        h1 = h1 * 103 + p[i][k];
        ++len;
        vh[m].first = len;
        vh[m].second = h1;
        ++m;
      }
    }
    sort(vh, vh + m);
    for (int j = 0; j < m;) {
      int e = j;
      while (e < m && vh[e] == vh[j]) ++e;
      if (l[i] == 0 || r[i] == 0) S2[n2++] = vh[j];
      if (r[i]) {
        if (l[i] == 0 && e - j > r[i]) S1[n1++] = vh[j];
        if (l[i] > 0 && e - j >= l[i] && e - j <= r[i]) S1[n1++] = vh[j];
      }
      j = e;
    }
    sort(S1, S1 + n1);
    sort(S2, S2 + n2);
    n1 = unique(S1, S1 + n1) - S1;
    n2 = unique(S2, S2 + n2) - S2;
    for (int j = 0; j < sz; ++j) {
      if (ok[j]) {
        bool ok2 = false;
        if (r[i] == 0) {
          if (!find2(n2, vh2[j])) ok2 = true;
        } else if (l[i] == 0) {
          if (!find2(n2, H))
            ok2 = true;
          else if (!find1(n1, vh2[j]))
            ok2 = true;
        } else {
          if (find1(n1, vh2[j])) ok2 = true;
        }
        ok[j] = ok2;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < sz; ++i)
    if (ok[i]) ++ans;
  printf("%d\n", ans);
  return 0;
}

```