#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x * f;
}
int d[11][50050 * 20], n, m, tot, las, l[11], r[11];
char s[50050];
int cc[50050], aa[50050 * 20];
long long ans;
struct node {
  int len, fa, ch[26];
} t[50050 * 20];
void Insert(int c, int x) {
  int p = las;
  if (t[p].ch[c] && t[t[p].ch[c]].len == t[p].len + 1)
    return (void)(++d[x][las = t[p].ch[c]]);
  int clone = 0, u = 0, flag = 0;
  if (t[p].ch[c])
    u = t[p].ch[c], flag = 1;
  else
    u = ++tot, t[u].len = t[p].len + 1;
  while (p && !t[p].ch[c]) {
    t[p].ch[c] = u, p = t[p].fa;
  }
  if (!p)
    t[u].fa = 1;
  else {
    int q = t[p].ch[c];
    if (t[q].len == t[p].len + 1)
      t[u].fa = q;
    else {
      clone = ++tot;
      t[clone] = t[q];
      t[clone].len = t[p].len + 1;
      while (p && t[p].ch[c] == q) {
        t[p].ch[c] = clone;
        p = t[p].fa;
      }
      t[u].fa = t[q].fa = clone;
    }
  }
  ++d[x][las = (flag ? clone : u)];
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  las = tot = 1;
  for (int i = 1; i <= n; ++i) {
    Insert(s[i] - 'a', 0);
  }
  m = read();
  for (int i = 1; i <= m; ++i) {
    las = 1;
    scanf("%s", s + 1);
    int qwq = strlen(s + 1);
    for (int j = 1; j <= qwq; ++j) {
      Insert(s[j] - 'a', i);
    }
    l[i] = read(), r[i] = read();
  }
  for (int i = 1; i <= tot; ++i) ++cc[t[i].len];
  for (int i = 1; i <= 50000; ++i) cc[i] += cc[i - 1];
  for (int i = 1; i <= tot; ++i) aa[cc[t[i].len]--] = i;
  for (int i = tot; i >= 2; --i) {
    int u = aa[i];
    bool ok = true;
    for (int j = 0; j <= m; ++j) {
      d[j][t[u].fa] += d[j][u];
    }
    for (int j = 1; j <= m; ++j) {
      if (d[j][u] < l[j] || d[j][u] > r[j]) {
        ok = false;
        break;
      }
    }
    if (ok && d[0][u]) ans += t[u].len - t[t[u].fa].len;
  }
  printf("%I64d\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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x * f;
}
int d[11][50050 * 20], n, m, tot, las, l[11], r[11];
char s[50050];
int cc[50050], aa[50050 * 20];
long long ans;
struct node {
  int len, fa, ch[26];
} t[50050 * 20];
void Insert(int c, int x) {
  int p = las;
  if (t[p].ch[c] && t[t[p].ch[c]].len == t[p].len + 1)
    return (void)(++d[x][las = t[p].ch[c]]);
  int clone = 0, u = 0, flag = 0;
  if (t[p].ch[c])
    u = t[p].ch[c], flag = 1;
  else
    u = ++tot, t[u].len = t[p].len + 1;
  while (p && !t[p].ch[c]) {
    t[p].ch[c] = u, p = t[p].fa;
  }
  if (!p)
    t[u].fa = 1;
  else {
    int q = t[p].ch[c];
    if (t[q].len == t[p].len + 1)
      t[u].fa = q;
    else {
      clone = ++tot;
      t[clone] = t[q];
      t[clone].len = t[p].len + 1;
      while (p && t[p].ch[c] == q) {
        t[p].ch[c] = clone;
        p = t[p].fa;
      }
      t[u].fa = t[q].fa = clone;
    }
  }
  ++d[x][las = (flag ? clone : u)];
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  las = tot = 1;
  for (int i = 1; i <= n; ++i) {
    Insert(s[i] - 'a', 0);
  }
  m = read();
  for (int i = 1; i <= m; ++i) {
    las = 1;
    scanf("%s", s + 1);
    int qwq = strlen(s + 1);
    for (int j = 1; j <= qwq; ++j) {
      Insert(s[j] - 'a', i);
    }
    l[i] = read(), r[i] = read();
  }
  for (int i = 1; i <= tot; ++i) ++cc[t[i].len];
  for (int i = 1; i <= 50000; ++i) cc[i] += cc[i - 1];
  for (int i = 1; i <= tot; ++i) aa[cc[t[i].len]--] = i;
  for (int i = tot; i >= 2; --i) {
    int u = aa[i];
    bool ok = true;
    for (int j = 0; j <= m; ++j) {
      d[j][t[u].fa] += d[j][u];
    }
    for (int j = 1; j <= m; ++j) {
      if (d[j][u] < l[j] || d[j][u] > r[j]) {
        ok = false;
        break;
      }
    }
    if (ok && d[0][u]) ans += t[u].len - t[t[u].fa].len;
  }
  printf("%I64d\n", ans);
  return 0;
}

```