#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int son[N][27], fail[N], sz[N][12], lst, tot, cnt, head[N], maxlen[N], n;
struct edge {
  int to, nxt;
} e[N * 2];
void add(int u, int v) {
  e[++cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void ins(int c) {
  if (son[lst][c]) {
    int p = lst, np = son[p][c];
    if (maxlen[p] + 1 == maxlen[np]) {
      lst = np;
      return;
    }
    int q = ++tot;
    maxlen[q] = maxlen[p] + 1;
    memcpy(son[q], son[np], sizeof son[np]);
    for (; p && son[p][c] == np; p = fail[p]) son[p][c] = q;
    fail[q] = fail[np];
    fail[np] = q;
    lst = q;
    return;
  }
  int p = lst, np = ++tot;
  maxlen[np] = maxlen[p] + 1;
  for (; p && !son[p][c]; p = fail[p]) son[p][c] = np;
  if (p == 0)
    fail[np] = 1;
  else {
    int q = son[p][c];
    if (maxlen[q] == maxlen[p] + 1)
      fail[np] = q;
    else {
      int nq = ++tot;
      fail[nq] = fail[q];
      memcpy(son[nq], son[q], sizeof son[nq]);
      maxlen[nq] = maxlen[p] + 1;
      for (; p && son[p][c] == q; p = fail[p]) son[p][c] = nq;
      fail[np] = fail[q] = nq;
    }
  }
  lst = np;
}
int l[20], r[20];
void dfs(int now) {
  for (int i = head[now]; i; i = e[i].nxt) {
    dfs(e[i].to);
    for (int j = 0; j <= n; j++) sz[now][j] += sz[e[i].to][j];
  }
}
bool check(int x) {
  bool flag = 1;
  for (int i = 1; i <= n; i++) flag &= (sz[x][i] >= l[i] && sz[x][i] <= r[i]);
  return flag;
}
int main() {
  string s;
  cin >> s;
  lst = tot = 1;
  for (int i = 0; i < s.size(); i++) ins(s[i] - 'a' + 1), sz[lst][0] = 1;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    lst = 1;
    cin >> s;
    for (int j = 0; j < s.size(); j++) ins(s[j] - 'a' + 1), sz[lst][i] = 1;
    cin >> l[i] >> r[i];
  }
  for (int i = 1; i <= tot; i++) add(fail[i], i);
  dfs(1);
  long long ans = 0;
  for (int i = 1; i <= tot; i++)
    if (sz[i][0] && check(i)) ans += maxlen[i] - maxlen[fail[i]];
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
const int N = 1e6 + 5;
int son[N][27], fail[N], sz[N][12], lst, tot, cnt, head[N], maxlen[N], n;
struct edge {
  int to, nxt;
} e[N * 2];
void add(int u, int v) {
  e[++cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void ins(int c) {
  if (son[lst][c]) {
    int p = lst, np = son[p][c];
    if (maxlen[p] + 1 == maxlen[np]) {
      lst = np;
      return;
    }
    int q = ++tot;
    maxlen[q] = maxlen[p] + 1;
    memcpy(son[q], son[np], sizeof son[np]);
    for (; p && son[p][c] == np; p = fail[p]) son[p][c] = q;
    fail[q] = fail[np];
    fail[np] = q;
    lst = q;
    return;
  }
  int p = lst, np = ++tot;
  maxlen[np] = maxlen[p] + 1;
  for (; p && !son[p][c]; p = fail[p]) son[p][c] = np;
  if (p == 0)
    fail[np] = 1;
  else {
    int q = son[p][c];
    if (maxlen[q] == maxlen[p] + 1)
      fail[np] = q;
    else {
      int nq = ++tot;
      fail[nq] = fail[q];
      memcpy(son[nq], son[q], sizeof son[nq]);
      maxlen[nq] = maxlen[p] + 1;
      for (; p && son[p][c] == q; p = fail[p]) son[p][c] = nq;
      fail[np] = fail[q] = nq;
    }
  }
  lst = np;
}
int l[20], r[20];
void dfs(int now) {
  for (int i = head[now]; i; i = e[i].nxt) {
    dfs(e[i].to);
    for (int j = 0; j <= n; j++) sz[now][j] += sz[e[i].to][j];
  }
}
bool check(int x) {
  bool flag = 1;
  for (int i = 1; i <= n; i++) flag &= (sz[x][i] >= l[i] && sz[x][i] <= r[i]);
  return flag;
}
int main() {
  string s;
  cin >> s;
  lst = tot = 1;
  for (int i = 0; i < s.size(); i++) ins(s[i] - 'a' + 1), sz[lst][0] = 1;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    lst = 1;
    cin >> s;
    for (int j = 0; j < s.size(); j++) ins(s[j] - 'a' + 1), sz[lst][i] = 1;
    cin >> l[i] >> r[i];
  }
  for (int i = 1; i <= tot; i++) add(fail[i], i);
  dfs(1);
  long long ans = 0;
  for (int i = 1; i <= tot; i++)
    if (sz[i][0] && check(i)) ans += maxlen[i] - maxlen[fail[i]];
  cout << ans;
  return 0;
}

```