#include <bits/stdc++.h>
using namespace std;
struct node {
  int len, link;
  int num[12];
  map<char, int> next;
};
node sam[(50100 * 24)];
int cnt, last;
void sam_init() {
  cnt = last = 0;
  sam[0].len = 0;
  sam[0].link = -1;
  cnt++;
}
void sam_extand(char c, int id) {
  int cur = cnt++;
  sam[cur].len = sam[last].len + 1;
  sam[cur].num[id]++;
  int p;
  for (p = last; p != -1 && !sam[p].next.count(c); p = sam[p].link)
    sam[p].next[c] = cur;
  if (p == -1)
    sam[cur].link = 0;
  else {
    int q = sam[p].next[c];
    if (sam[p].len + 1 == sam[q].len)
      sam[cur].link = q;
    else {
      int clone = cnt++;
      sam[clone].len = sam[p].len + 1;
      sam[clone].next = sam[q].next;
      sam[clone].link = sam[q].link;
      for (; p != -1 && sam[p].next[c] == q; p = sam[p].link)
        sam[p].next[c] = clone;
      sam[q].link = sam[cur].link = clone;
    }
  }
  last = cur;
}
char s[50100];
char t[15][50100];
int lb[15], rb[15];
queue<int> q;
int fa[(50100 * 24)];
int n;
bool check(int x) {
  if (sam[x].num[0] == 0) return false;
  for (int i = 1; i <= n; i++)
    if (sam[x].num[i] < lb[i] || sam[x].num[i] > rb[i]) return false;
  return true;
}
int main() {
  scanf("%s", s + 1);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", t[i] + 1);
    scanf("%d%d", lb + i, rb + i);
  }
  sam_init();
  for (int i = 1, j = strlen(s + 1); i <= j; i++) sam_extand(s[i], 0);
  for (int i = 1; i <= n; i++) {
    sam_extand('#', i);
    for (int j = 1, k = strlen(t[i] + 1); j <= k; j++) sam_extand(t[i][j], i);
  }
  for (int i = 1; i < cnt; i++)
    if (sam[i].link != -1) fa[sam[i].link]++;
  for (int i = 1; i < cnt; i++)
    if (!fa[i]) q.push(i);
  int ans = 0;
  while (!q.empty()) {
    int x = q.front();
    int y = sam[x].link;
    if (x == 0) break;
    q.pop();
    if (check(x)) ans += sam[x].len - sam[y].len;
    for (int i = 0; i <= n; i++) sam[y].num[i] += sam[x].num[i];
    fa[y]--;
    if (fa[y] == 0) q.push(y);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
struct node {
  int len, link;
  int num[12];
  map<char, int> next;
};
node sam[(50100 * 24)];
int cnt, last;
void sam_init() {
  cnt = last = 0;
  sam[0].len = 0;
  sam[0].link = -1;
  cnt++;
}
void sam_extand(char c, int id) {
  int cur = cnt++;
  sam[cur].len = sam[last].len + 1;
  sam[cur].num[id]++;
  int p;
  for (p = last; p != -1 && !sam[p].next.count(c); p = sam[p].link)
    sam[p].next[c] = cur;
  if (p == -1)
    sam[cur].link = 0;
  else {
    int q = sam[p].next[c];
    if (sam[p].len + 1 == sam[q].len)
      sam[cur].link = q;
    else {
      int clone = cnt++;
      sam[clone].len = sam[p].len + 1;
      sam[clone].next = sam[q].next;
      sam[clone].link = sam[q].link;
      for (; p != -1 && sam[p].next[c] == q; p = sam[p].link)
        sam[p].next[c] = clone;
      sam[q].link = sam[cur].link = clone;
    }
  }
  last = cur;
}
char s[50100];
char t[15][50100];
int lb[15], rb[15];
queue<int> q;
int fa[(50100 * 24)];
int n;
bool check(int x) {
  if (sam[x].num[0] == 0) return false;
  for (int i = 1; i <= n; i++)
    if (sam[x].num[i] < lb[i] || sam[x].num[i] > rb[i]) return false;
  return true;
}
int main() {
  scanf("%s", s + 1);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", t[i] + 1);
    scanf("%d%d", lb + i, rb + i);
  }
  sam_init();
  for (int i = 1, j = strlen(s + 1); i <= j; i++) sam_extand(s[i], 0);
  for (int i = 1; i <= n; i++) {
    sam_extand('#', i);
    for (int j = 1, k = strlen(t[i] + 1); j <= k; j++) sam_extand(t[i][j], i);
  }
  for (int i = 1; i < cnt; i++)
    if (sam[i].link != -1) fa[sam[i].link]++;
  for (int i = 1; i < cnt; i++)
    if (!fa[i]) q.push(i);
  int ans = 0;
  while (!q.empty()) {
    int x = q.front();
    int y = sam[x].link;
    if (x == 0) break;
    q.pop();
    if (check(x)) ans += sam[x].len - sam[y].len;
    for (int i = 0; i <= n; i++) sam[y].num[i] += sam[x].num[i];
    fa[y]--;
    if (fa[y] == 0) q.push(y);
  }
  printf("%d\n", ans);
  return 0;
}

```