#include <bits/stdc++.h>
using namespace std;
struct node {
  int son[26], fa, right, step;
} a[100010];
int lm[50010], rm[50010], n, len, ans, sz, last, in[100010], q[100010],
    c[100010], d[100010], e[100010];
char s[50010], p[50010];
void add(int x) {
  int p = last, np = ++sz;
  a[np].step = a[p].step + 1;
  a[np].right = 1;
  for (; p && !a[p].son[x]; p = a[p].fa) a[p].son[x] = np;
  if (!p)
    a[np].fa = 1;
  else if (a[a[p].son[x]].step == a[p].step + 1)
    a[np].fa = a[p].son[x];
  else {
    int nq = ++sz, q = a[p].son[x];
    a[nq] = a[q];
    a[nq].right = 0;
    a[q].fa = a[np].fa = nq;
    a[nq].step = a[p].step + 1;
    for (; p && a[p].son[x] == q; p = a[p].fa) a[p].son[x] = nq;
  }
  last = np;
}
void build() {
  memset(a, 0, sizeof(a));
  last = sz = 1;
  int slen = strlen(p + 1);
  for (int i = 1; i <= slen; i++) add(p[i] - 'a');
  for (int i = 1; i <= sz; i++) in[a[i].fa]++;
  int l = 0, r = 0;
  for (int i = 1; i <= sz; i++)
    if (!in[i]) q[++r] = i;
  while (l < r) {
    int x = q[++l];
    a[a[x].fa].right += a[x].right;
    if (!--in[a[x].fa]) q[++r] = a[x].fa;
  }
}
void work(int lim, int type) {
  int now = 1, lp = 0;
  for (int i = 1; i <= len; i++) {
    if (a[now].son[s[i] - 'a'])
      ++lp, now = a[now].son[s[i] - 'a'];
    else {
      for (; now && !a[now].son[s[i] - 'a']; now = a[now].fa)
        ;
      if (!now)
        now = 1, lp = 0;
      else
        lp = a[now].step + 1, now = a[now].son[s[i] - 'a'];
    }
    while (lp && a[now].right < lim) now = a[now].fa, lp = a[now].step;
    if (type)
      rm[i] = min(rm[i], i - lp);
    else
      lm[i] = max(lm[i], i - lp + 1);
  }
}
int main() {
  scanf("%s%d", s + 1, &n);
  len = strlen(s + 1);
  for (int i = 1; i <= len; i++) rm[i] = i, lm[i] = 1;
  for (int i = 1; i <= n; i++) {
    int l, r;
    scanf("%s%d%d", p + 1, &l, &r);
    build();
    if (l) work(l, 0);
    work(r + 1, 1);
  }
  memset(a, 0, sizeof(a));
  last = sz = 1;
  for (int i = 1; i <= len; i++) add(s[i] - 'a');
  for (int i = 1; i <= sz; i++) c[i] = a[i].step, d[i] = a[a[i].fa].step + 1;
  int now = 1, lp = 0;
  for (int i = 1; i <= len; i++) {
    if (a[now].son[s[i] - 'a'])
      ++lp, now = a[now].son[s[i] - 'a'];
    else {
      for (; now && !a[now].son[s[i] - 'a']; now = a[now].fa)
        ;
      if (!now)
        now = 1, lp = 0;
      else
        lp = a[now].step + 1, now = a[now].son[s[i] - 'a'];
    }
    while (i - lp + 1 < lm[i]) {
      if (i - a[a[now].fa].step + 1 <= lm[i])
        now = a[now].fa, lp = a[now].step;
      else {
        lp = i - lm[i] + 1;
        break;
      }
    }
    c[now] = min(c[now], lp);
    e[now] |= 1;
  }
  now = 1, lp = 0;
  for (int i = 1; i <= len; i++) {
    if (a[now].son[s[i] - 'a'])
      ++lp, now = a[now].son[s[i] - 'a'];
    else {
      for (; now && !a[now].son[s[i] - 'a']; now = a[now].fa)
        ;
      if (!now)
        now = 1, lp = 0;
      else
        lp = a[now].step + 1, now = a[now].son[s[i] - 'a'];
    }
    while (now && i - lp + 1 <= rm[i]) {
      if (i - a[a[now].fa].step + 1 <= rm[i] + 1)
        now = a[now].fa, lp = a[now].step;
      else {
        lp = i - rm[i];
        break;
      }
    }
    d[now] = max(d[now], lp + 1);
    e[now] |= 1;
    e[a[now].fa] |= -1;
    if (!now) now = 1;
  }
  for (int i = 1; i <= sz; i++) in[a[i].fa]++;
  int l = 0, r = 0;
  for (int i = 1; i <= sz; i++)
    if (!in[i]) q[++r] = i;
  while (l < r) {
    int x = q[++l];
    a[a[x].fa].right += a[x].right;
    e[a[x].fa] |= e[x];
    if (!--in[a[x].fa]) q[++r] = a[x].fa;
  }
  for (int i = 2; i <= sz; i++)
    if (e[i] == 1 && c[i] >= d[i]) ans += c[i] - d[i] + 1;
  printf("%d\n", ans);
}

### 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 son[26], fa, right, step;
} a[100010];
int lm[50010], rm[50010], n, len, ans, sz, last, in[100010], q[100010],
    c[100010], d[100010], e[100010];
char s[50010], p[50010];
void add(int x) {
  int p = last, np = ++sz;
  a[np].step = a[p].step + 1;
  a[np].right = 1;
  for (; p && !a[p].son[x]; p = a[p].fa) a[p].son[x] = np;
  if (!p)
    a[np].fa = 1;
  else if (a[a[p].son[x]].step == a[p].step + 1)
    a[np].fa = a[p].son[x];
  else {
    int nq = ++sz, q = a[p].son[x];
    a[nq] = a[q];
    a[nq].right = 0;
    a[q].fa = a[np].fa = nq;
    a[nq].step = a[p].step + 1;
    for (; p && a[p].son[x] == q; p = a[p].fa) a[p].son[x] = nq;
  }
  last = np;
}
void build() {
  memset(a, 0, sizeof(a));
  last = sz = 1;
  int slen = strlen(p + 1);
  for (int i = 1; i <= slen; i++) add(p[i] - 'a');
  for (int i = 1; i <= sz; i++) in[a[i].fa]++;
  int l = 0, r = 0;
  for (int i = 1; i <= sz; i++)
    if (!in[i]) q[++r] = i;
  while (l < r) {
    int x = q[++l];
    a[a[x].fa].right += a[x].right;
    if (!--in[a[x].fa]) q[++r] = a[x].fa;
  }
}
void work(int lim, int type) {
  int now = 1, lp = 0;
  for (int i = 1; i <= len; i++) {
    if (a[now].son[s[i] - 'a'])
      ++lp, now = a[now].son[s[i] - 'a'];
    else {
      for (; now && !a[now].son[s[i] - 'a']; now = a[now].fa)
        ;
      if (!now)
        now = 1, lp = 0;
      else
        lp = a[now].step + 1, now = a[now].son[s[i] - 'a'];
    }
    while (lp && a[now].right < lim) now = a[now].fa, lp = a[now].step;
    if (type)
      rm[i] = min(rm[i], i - lp);
    else
      lm[i] = max(lm[i], i - lp + 1);
  }
}
int main() {
  scanf("%s%d", s + 1, &n);
  len = strlen(s + 1);
  for (int i = 1; i <= len; i++) rm[i] = i, lm[i] = 1;
  for (int i = 1; i <= n; i++) {
    int l, r;
    scanf("%s%d%d", p + 1, &l, &r);
    build();
    if (l) work(l, 0);
    work(r + 1, 1);
  }
  memset(a, 0, sizeof(a));
  last = sz = 1;
  for (int i = 1; i <= len; i++) add(s[i] - 'a');
  for (int i = 1; i <= sz; i++) c[i] = a[i].step, d[i] = a[a[i].fa].step + 1;
  int now = 1, lp = 0;
  for (int i = 1; i <= len; i++) {
    if (a[now].son[s[i] - 'a'])
      ++lp, now = a[now].son[s[i] - 'a'];
    else {
      for (; now && !a[now].son[s[i] - 'a']; now = a[now].fa)
        ;
      if (!now)
        now = 1, lp = 0;
      else
        lp = a[now].step + 1, now = a[now].son[s[i] - 'a'];
    }
    while (i - lp + 1 < lm[i]) {
      if (i - a[a[now].fa].step + 1 <= lm[i])
        now = a[now].fa, lp = a[now].step;
      else {
        lp = i - lm[i] + 1;
        break;
      }
    }
    c[now] = min(c[now], lp);
    e[now] |= 1;
  }
  now = 1, lp = 0;
  for (int i = 1; i <= len; i++) {
    if (a[now].son[s[i] - 'a'])
      ++lp, now = a[now].son[s[i] - 'a'];
    else {
      for (; now && !a[now].son[s[i] - 'a']; now = a[now].fa)
        ;
      if (!now)
        now = 1, lp = 0;
      else
        lp = a[now].step + 1, now = a[now].son[s[i] - 'a'];
    }
    while (now && i - lp + 1 <= rm[i]) {
      if (i - a[a[now].fa].step + 1 <= rm[i] + 1)
        now = a[now].fa, lp = a[now].step;
      else {
        lp = i - rm[i];
        break;
      }
    }
    d[now] = max(d[now], lp + 1);
    e[now] |= 1;
    e[a[now].fa] |= -1;
    if (!now) now = 1;
  }
  for (int i = 1; i <= sz; i++) in[a[i].fa]++;
  int l = 0, r = 0;
  for (int i = 1; i <= sz; i++)
    if (!in[i]) q[++r] = i;
  while (l < r) {
    int x = q[++l];
    a[a[x].fa].right += a[x].right;
    e[a[x].fa] |= e[x];
    if (!--in[a[x].fa]) q[++r] = a[x].fa;
  }
  for (int i = 2; i <= sz; i++)
    if (e[i] == 1 && c[i] >= d[i]) ans += c[i] - d[i] + 1;
  printf("%d\n", ans);
}

```