#include <bits/stdc++.h>
using namespace std;
char s[50010];
int n, m;
char t[10][50010];
int tn[10], tl[10], tr[10];
int p[50010];
int ps[50010], pt[10][50010];
int rs[50010], rt[10][50010];
int lb;
int *pw;
int *rw;
int hhx(int l, int r) { return ps[r] - ps[l] * p[r - l]; }
int hhy(int l, int r) { return pw[r] - pw[l] * p[r - l]; }
int lcps(int x, int y) {
  int l = 1, r = min(n - x, n - y);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (hhx(x, x + mid) == hhx(y, y + mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
int lcpt(int x, int y) {
  int l = 1, r = min(tn[lb] - x, tn[lb] - y);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (hhy(x, x + mid) == hhy(y, y + mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
bool cmps(int x, int y) {
  int c = lcps(x, y);
  return s[x + c] < s[y + c];
}
bool cmpt(int x, int y) {
  int c = lcpt(x, y);
  return t[lb][x + c] < t[lb][y + c];
}
int lcpst(int x, int y, int z) {
  int l = 1, r = min(y - x, tn[lb] - z);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (hhx(x, x + mid) == hhy(z, z + mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
int getl(int l, int r, int b) {
  lb = b;
  pw = pt[lb];
  rw = rt[lb];
  int le1 = 0, ri1 = tn[lb] - 1;
  while (le1 <= ri1) {
    int mid = (le1 + ri1) / 2;
    int c = lcpst(l, r, rw[mid]);
    if (r == l + c)
      ri1 = mid - 1;
    else if (s[l + c] < t[lb][rw[mid] + c])
      ri1 = mid - 1;
    else
      le1 = mid + 1;
  }
  int le2 = 0, ri2 = tn[lb] - 1;
  while (le2 <= ri2) {
    int mid = (le2 + ri2) / 2;
    int c = lcpst(l, r, rw[mid]);
    if (r == l + c)
      le2 = mid + 1;
    else if (s[l + c] > t[lb][rw[mid] + c])
      le2 = mid + 1;
    else
      ri2 = mid - 1;
  }
  return ri2 - le1 + 1;
}
bool F;
bool cmp1(int x, int y) {
  bool F0 = 0, F1 = 0;
  for (int i = 0; i < m; i++) {
    int l = getl(x, y, i);
    if (l < tl[i]) F0 = 1;
    if (l > tr[i]) F1 = 1;
    if (F0 && F1) {
      F = 1;
      return 0;
    }
  }
  if (!F1) return 1;
  return 0;
}
bool cmp2(int x, int y) {
  bool F0 = 0, F1 = 0;
  for (int i = 0; i < m; i++) {
    int l = getl(x, y, i);
    if (l < tl[i]) F0 = 1;
    if (l > tr[i]) F1 = 1;
    if (F0 && F1) {
      F = 1;
      return 0;
    }
  }
  if (!F0) return 1;
  return 0;
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  scanf("%d", &m);
  for (int i = 0; i < m; i++)
    scanf("%s%d%d", t[i], tl + i, tr + i), tn[i] = strlen(t[i]);
  p[0] = 1;
  for (int i = 1; i < 50010; i++) p[i] = p[i - 1] * 1000000007;
  ps[0] = 0;
  for (int i = 0; i < n; i++) ps[i + 1] = ps[i] * 1000000007 + s[i];
  for (int i = 0; i < m; i++) {
    int *pw = pt[i];
    char *st = t[i];
    pw[0] = 0;
    for (int j = 0; j < tn[i]; j++) pw[j + 1] = pw[j] * 1000000007 + st[j];
  }
  for (int i = 0; i < n; i++) rs[i] = i;
  sort(rs, rs + n, cmps);
  for (int i = 0; i < m; i++) {
    int *rw = rt[i];
    for (int j = 0; j < tn[i]; j++) rw[j] = j;
    pw = pt[i];
    lb = i;
    sort(rw, rw + tn[i], cmpt);
  }
  int S = 0;
  for (int i = 0; i < n; i++) {
    int le = 0;
    if (i > 0) le = lcps(rs[i], rs[i - 1]);
    int ri = n - rs[i];
    F = 0;
    int le1 = le + 1, ri1 = ri;
    while (le1 <= ri1) {
      int mid = (le1 + ri1) / 2;
      if (cmp1(rs[i], rs[i] + mid))
        ri1 = mid - 1;
      else
        le1 = mid + 1;
      if (F) break;
    }
    if (F) continue;
    int le2 = le + 1, ri2 = ri;
    while (le2 <= ri2) {
      int mid = (le2 + ri2) / 2;
      if (cmp2(rs[i], rs[i] + mid))
        le2 = mid + 1;
      else
        ri2 = mid - 1;
      if (F) break;
    }
    if (F) continue;
    S += max(0, ri2 - le1 + 1);
  }
  printf("%d\n", S);
  return 0;
}

### Prompt

Please create a solution in CPP 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 s[50010];
int n, m;
char t[10][50010];
int tn[10], tl[10], tr[10];
int p[50010];
int ps[50010], pt[10][50010];
int rs[50010], rt[10][50010];
int lb;
int *pw;
int *rw;
int hhx(int l, int r) { return ps[r] - ps[l] * p[r - l]; }
int hhy(int l, int r) { return pw[r] - pw[l] * p[r - l]; }
int lcps(int x, int y) {
  int l = 1, r = min(n - x, n - y);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (hhx(x, x + mid) == hhx(y, y + mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
int lcpt(int x, int y) {
  int l = 1, r = min(tn[lb] - x, tn[lb] - y);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (hhy(x, x + mid) == hhy(y, y + mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
bool cmps(int x, int y) {
  int c = lcps(x, y);
  return s[x + c] < s[y + c];
}
bool cmpt(int x, int y) {
  int c = lcpt(x, y);
  return t[lb][x + c] < t[lb][y + c];
}
int lcpst(int x, int y, int z) {
  int l = 1, r = min(y - x, tn[lb] - z);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (hhx(x, x + mid) == hhy(z, z + mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
int getl(int l, int r, int b) {
  lb = b;
  pw = pt[lb];
  rw = rt[lb];
  int le1 = 0, ri1 = tn[lb] - 1;
  while (le1 <= ri1) {
    int mid = (le1 + ri1) / 2;
    int c = lcpst(l, r, rw[mid]);
    if (r == l + c)
      ri1 = mid - 1;
    else if (s[l + c] < t[lb][rw[mid] + c])
      ri1 = mid - 1;
    else
      le1 = mid + 1;
  }
  int le2 = 0, ri2 = tn[lb] - 1;
  while (le2 <= ri2) {
    int mid = (le2 + ri2) / 2;
    int c = lcpst(l, r, rw[mid]);
    if (r == l + c)
      le2 = mid + 1;
    else if (s[l + c] > t[lb][rw[mid] + c])
      le2 = mid + 1;
    else
      ri2 = mid - 1;
  }
  return ri2 - le1 + 1;
}
bool F;
bool cmp1(int x, int y) {
  bool F0 = 0, F1 = 0;
  for (int i = 0; i < m; i++) {
    int l = getl(x, y, i);
    if (l < tl[i]) F0 = 1;
    if (l > tr[i]) F1 = 1;
    if (F0 && F1) {
      F = 1;
      return 0;
    }
  }
  if (!F1) return 1;
  return 0;
}
bool cmp2(int x, int y) {
  bool F0 = 0, F1 = 0;
  for (int i = 0; i < m; i++) {
    int l = getl(x, y, i);
    if (l < tl[i]) F0 = 1;
    if (l > tr[i]) F1 = 1;
    if (F0 && F1) {
      F = 1;
      return 0;
    }
  }
  if (!F0) return 1;
  return 0;
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  scanf("%d", &m);
  for (int i = 0; i < m; i++)
    scanf("%s%d%d", t[i], tl + i, tr + i), tn[i] = strlen(t[i]);
  p[0] = 1;
  for (int i = 1; i < 50010; i++) p[i] = p[i - 1] * 1000000007;
  ps[0] = 0;
  for (int i = 0; i < n; i++) ps[i + 1] = ps[i] * 1000000007 + s[i];
  for (int i = 0; i < m; i++) {
    int *pw = pt[i];
    char *st = t[i];
    pw[0] = 0;
    for (int j = 0; j < tn[i]; j++) pw[j + 1] = pw[j] * 1000000007 + st[j];
  }
  for (int i = 0; i < n; i++) rs[i] = i;
  sort(rs, rs + n, cmps);
  for (int i = 0; i < m; i++) {
    int *rw = rt[i];
    for (int j = 0; j < tn[i]; j++) rw[j] = j;
    pw = pt[i];
    lb = i;
    sort(rw, rw + tn[i], cmpt);
  }
  int S = 0;
  for (int i = 0; i < n; i++) {
    int le = 0;
    if (i > 0) le = lcps(rs[i], rs[i - 1]);
    int ri = n - rs[i];
    F = 0;
    int le1 = le + 1, ri1 = ri;
    while (le1 <= ri1) {
      int mid = (le1 + ri1) / 2;
      if (cmp1(rs[i], rs[i] + mid))
        ri1 = mid - 1;
      else
        le1 = mid + 1;
      if (F) break;
    }
    if (F) continue;
    int le2 = le + 1, ri2 = ri;
    while (le2 <= ri2) {
      int mid = (le2 + ri2) / 2;
      if (cmp2(rs[i], rs[i] + mid))
        le2 = mid + 1;
      else
        ri2 = mid - 1;
      if (F) break;
    }
    if (F) continue;
    S += max(0, ri2 - le1 + 1);
  }
  printf("%d\n", S);
  return 0;
}

```