#include <bits/stdc++.h>
using namespace std;
char str[12][409600];
int n;
int ch[2 * 409600][26], pre[2 * 409600], len[2 * 409600];
int siz, fin;
int ll[15], L[15], R[15];
int s[3 * 409600][12];
struct edge {
  int to;
  int nxt;
} nd[3 * 409600];
int head[3 * 409600], cnt;
int ans;
void insert(int c) {
  int p, np, q, nq;
  if (ch[fin][c]) {
    p = fin;
    q = ch[fin][c];
    if (len[q] == len[p] + 1) {
      fin = q;
      return;
    } else {
      nq = ++siz;
      memcpy(ch[nq], ch[q], sizeof(ch[nq]));
      len[nq] = len[p] + 1;
      pre[nq] = pre[q];
      pre[q] = nq;
      while (p && ch[p][c] == q) {
        ch[p][c] = nq;
        p = pre[p];
      }
      fin = nq;
      return;
    }
  }
  p = fin;
  np = fin = ++siz;
  len[np] = len[p] + 1;
  while (p && !ch[p][c]) {
    ch[p][c] = np;
    p = pre[p];
  }
  if (!p) {
    pre[np] = 1;
  } else {
    q = ch[p][c];
    if (len[q] == len[p] + 1) {
      pre[np] = q;
      return;
    } else {
      nq = ++siz;
      memcpy(ch[nq], ch[q], sizeof(ch[nq]));
      len[nq] = len[p] + 1;
      pre[nq] = pre[q];
      pre[np] = pre[q] = nq;
      while (p && ch[p][c] == q) {
        ch[p][c] = nq;
        p = pre[p];
      }
    }
  }
  return;
}
void init() {
  memset(head, -1, sizeof(head));
  cnt = fin = siz = 1;
  return;
}
void add(int x, int y) {
  nd[cnt].to = y;
  nd[cnt].nxt = head[x];
  head[x] = cnt++;
  return;
}
void dfs(int x) {
  for (int i = head[x]; i != -1; i = nd[i].nxt) {
    int t = nd[i].to;
    dfs(t);
    for (int j = 0; j <= n; j++) {
      s[x][j] += s[t][j];
    }
  }
  return;
}
int main() {
  init();
  scanf("%s%d", str[0] + 1, &n);
  ll[0] = strlen(str[0] + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%s%d%d", str[i] + 1, &L[i], &R[i]);
    ll[i] = strlen(str[i] + 1);
  }
  for (int i = 0; i <= n; i++) {
    fin = 1;
    for (int j = 1; j <= ll[i]; j++) {
      insert(str[i][j] - 'a');
      s[fin][i]++;
    }
  }
  for (int i = 2; i <= siz; i++) {
    add(pre[i], i);
  }
  dfs(1);
  for (int i = 2; i <= siz; i++) {
    if (s[i][0]) {
      int j;
      for (j = 1; j <= n; j++) {
        if (s[i][j] < L[j] || s[i][j] > R[j]) {
          break;
        }
      }
      if (j > n) {
        ans += (len[i] - len[pre[i]]);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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;
char str[12][409600];
int n;
int ch[2 * 409600][26], pre[2 * 409600], len[2 * 409600];
int siz, fin;
int ll[15], L[15], R[15];
int s[3 * 409600][12];
struct edge {
  int to;
  int nxt;
} nd[3 * 409600];
int head[3 * 409600], cnt;
int ans;
void insert(int c) {
  int p, np, q, nq;
  if (ch[fin][c]) {
    p = fin;
    q = ch[fin][c];
    if (len[q] == len[p] + 1) {
      fin = q;
      return;
    } else {
      nq = ++siz;
      memcpy(ch[nq], ch[q], sizeof(ch[nq]));
      len[nq] = len[p] + 1;
      pre[nq] = pre[q];
      pre[q] = nq;
      while (p && ch[p][c] == q) {
        ch[p][c] = nq;
        p = pre[p];
      }
      fin = nq;
      return;
    }
  }
  p = fin;
  np = fin = ++siz;
  len[np] = len[p] + 1;
  while (p && !ch[p][c]) {
    ch[p][c] = np;
    p = pre[p];
  }
  if (!p) {
    pre[np] = 1;
  } else {
    q = ch[p][c];
    if (len[q] == len[p] + 1) {
      pre[np] = q;
      return;
    } else {
      nq = ++siz;
      memcpy(ch[nq], ch[q], sizeof(ch[nq]));
      len[nq] = len[p] + 1;
      pre[nq] = pre[q];
      pre[np] = pre[q] = nq;
      while (p && ch[p][c] == q) {
        ch[p][c] = nq;
        p = pre[p];
      }
    }
  }
  return;
}
void init() {
  memset(head, -1, sizeof(head));
  cnt = fin = siz = 1;
  return;
}
void add(int x, int y) {
  nd[cnt].to = y;
  nd[cnt].nxt = head[x];
  head[x] = cnt++;
  return;
}
void dfs(int x) {
  for (int i = head[x]; i != -1; i = nd[i].nxt) {
    int t = nd[i].to;
    dfs(t);
    for (int j = 0; j <= n; j++) {
      s[x][j] += s[t][j];
    }
  }
  return;
}
int main() {
  init();
  scanf("%s%d", str[0] + 1, &n);
  ll[0] = strlen(str[0] + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%s%d%d", str[i] + 1, &L[i], &R[i]);
    ll[i] = strlen(str[i] + 1);
  }
  for (int i = 0; i <= n; i++) {
    fin = 1;
    for (int j = 1; j <= ll[i]; j++) {
      insert(str[i][j] - 'a');
      s[fin][i]++;
    }
  }
  for (int i = 2; i <= siz; i++) {
    add(pre[i], i);
  }
  dfs(1);
  for (int i = 2; i <= siz; i++) {
    if (s[i][0]) {
      int j;
      for (j = 1; j <= n; j++) {
        if (s[i][j] < L[j] || s[i][j] > R[j]) {
          break;
        }
      }
      if (j > n) {
        ans += (len[i] - len[pre[i]]);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```