#include <bits/stdc++.h>
using namespace std;
struct node {
  int to[27];
  int par, l, r[12];
} ns[1010101];
int root = 1, nn = 1, last = 1;
void add(int x, int lb) {
  int p = last;
  int np = ++nn;
  last = np;
  ns[np].l = ns[p].l + 1;
  ns[np].r[lb] = 1;
  while (p && !ns[p].to[x]) ns[p].to[x] = np, p = ns[p].par;
  if (!p)
    ns[np].par = root;
  else {
    int q = ns[p].to[x];
    if (ns[q].l == ns[p].l + 1)
      ns[np].par = q;
    else {
      int nq = ++nn;
      memcpy(ns[nq].to, ns[q].to, sizeof(ns[q].to));
      ns[nq].l = ns[p].l + 1;
      ns[nq].par = ns[q].par;
      ns[q].par = ns[np].par = nq;
      while (p && ns[p].to[x] == q) ns[p].to[x] = nq, p = ns[p].par;
    }
  }
}
void add(char *a, int l, int lb) {
  for (int i = 0; i < l; i++) add(a[i] - 'a', lb);
  add(26, 11);
}
int rd[1010101];
int l[11], r[11];
char bf[50505];
int main() {
  int n;
  scanf("%s%d", bf, &n);
  add(bf, strlen(bf), 0);
  for (int i = 1; i <= n; i++) {
    scanf("%s%d%d", bf, l + i, r + i);
    add(bf, strlen(bf), i);
  }
  for (int i = 2; i <= nn; i++) rd[ns[i].par]++;
  queue<int> q;
  for (int i = 2; i <= nn; i++)
    if (!rd[i]) q.push(i);
  long long ans = 0;
  while (!q.empty()) {
    int nw = q.front();
    q.pop();
    if (nw == root) continue;
    if (!--rd[ns[nw].par]) q.push(ns[nw].par);
    for (int i = 0; i <= n; i++) ns[ns[nw].par].r[i] += ns[nw].r[i];
    bool fl = ns[nw].r[0];
    for (int i = 1; fl && i <= n; i++)
      if (ns[nw].r[i] > r[i] || ns[nw].r[i] < l[i]) fl = 0;
    if (fl) ans += ns[nw].l - ns[ns[nw].par].l;
  }
  printf("%I64d", ans);
  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;
struct node {
  int to[27];
  int par, l, r[12];
} ns[1010101];
int root = 1, nn = 1, last = 1;
void add(int x, int lb) {
  int p = last;
  int np = ++nn;
  last = np;
  ns[np].l = ns[p].l + 1;
  ns[np].r[lb] = 1;
  while (p && !ns[p].to[x]) ns[p].to[x] = np, p = ns[p].par;
  if (!p)
    ns[np].par = root;
  else {
    int q = ns[p].to[x];
    if (ns[q].l == ns[p].l + 1)
      ns[np].par = q;
    else {
      int nq = ++nn;
      memcpy(ns[nq].to, ns[q].to, sizeof(ns[q].to));
      ns[nq].l = ns[p].l + 1;
      ns[nq].par = ns[q].par;
      ns[q].par = ns[np].par = nq;
      while (p && ns[p].to[x] == q) ns[p].to[x] = nq, p = ns[p].par;
    }
  }
}
void add(char *a, int l, int lb) {
  for (int i = 0; i < l; i++) add(a[i] - 'a', lb);
  add(26, 11);
}
int rd[1010101];
int l[11], r[11];
char bf[50505];
int main() {
  int n;
  scanf("%s%d", bf, &n);
  add(bf, strlen(bf), 0);
  for (int i = 1; i <= n; i++) {
    scanf("%s%d%d", bf, l + i, r + i);
    add(bf, strlen(bf), i);
  }
  for (int i = 2; i <= nn; i++) rd[ns[i].par]++;
  queue<int> q;
  for (int i = 2; i <= nn; i++)
    if (!rd[i]) q.push(i);
  long long ans = 0;
  while (!q.empty()) {
    int nw = q.front();
    q.pop();
    if (nw == root) continue;
    if (!--rd[ns[nw].par]) q.push(ns[nw].par);
    for (int i = 0; i <= n; i++) ns[ns[nw].par].r[i] += ns[nw].r[i];
    bool fl = ns[nw].r[0];
    for (int i = 1; fl && i <= n; i++)
      if (ns[nw].r[i] > r[i] || ns[nw].r[i] < l[i]) fl = 0;
    if (fl) ans += ns[nw].l - ns[ns[nw].par].l;
  }
  printf("%I64d", ans);
  return 0;
}

```