#include <bits/stdc++.h>
using namespace std;
char pp[2005][2005];
int Len[2005];
int L[2005], R[2005];
char str[2005];
int pre[2000005][26] = {0}, allc, m;
struct node {
  int cnt;
  int Q[2005];
} turn[2][10];
int down(int cur, char c) {
  int nxt = cur ^ 1;
  int i, j, k;
  int res = 1;
  for (i = 0; i < m; i++) {
    turn[nxt][i].cnt = 0;
    for (j = 0; j < turn[cur][i].cnt; j++) {
      int pre = turn[cur][i].Q[j];
      if (pre < Len[i] - 1 && pp[i][pre + 1] == c) {
        turn[nxt][i].Q[turn[nxt][i].cnt++] = pre + 1;
      }
    }
    int cnt = turn[nxt][i].cnt;
    if (cnt > R[i] || cnt < L[i]) res = 0;
  }
  return res;
}
int main() {
  int n, i, j, k;
  scanf("%s", str);
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    scanf("%s %d %d", pp[i], &L[i], &R[i]);
    Len[i] = strlen(pp[i]);
  }
  allc = 1;
  n = strlen(str);
  int ans = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      turn[0][j].cnt = 0;
      for (k = 0; k < Len[j]; k++) turn[0][j].Q[turn[0][j].cnt++] = k - 1;
    }
    int cur = 0;
    for (j = i; j < n; j++) {
      int flag = down((j - i) & 1, str[j]);
      if (pre[cur][str[j] - 'a'] == 0) {
        if (flag == 1) ans++;
        pre[cur][str[j] - 'a'] = allc++;
      }
      cur = pre[cur][str[j] - 'a'];
    }
  }
  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 pp[2005][2005];
int Len[2005];
int L[2005], R[2005];
char str[2005];
int pre[2000005][26] = {0}, allc, m;
struct node {
  int cnt;
  int Q[2005];
} turn[2][10];
int down(int cur, char c) {
  int nxt = cur ^ 1;
  int i, j, k;
  int res = 1;
  for (i = 0; i < m; i++) {
    turn[nxt][i].cnt = 0;
    for (j = 0; j < turn[cur][i].cnt; j++) {
      int pre = turn[cur][i].Q[j];
      if (pre < Len[i] - 1 && pp[i][pre + 1] == c) {
        turn[nxt][i].Q[turn[nxt][i].cnt++] = pre + 1;
      }
    }
    int cnt = turn[nxt][i].cnt;
    if (cnt > R[i] || cnt < L[i]) res = 0;
  }
  return res;
}
int main() {
  int n, i, j, k;
  scanf("%s", str);
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    scanf("%s %d %d", pp[i], &L[i], &R[i]);
    Len[i] = strlen(pp[i]);
  }
  allc = 1;
  n = strlen(str);
  int ans = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      turn[0][j].cnt = 0;
      for (k = 0; k < Len[j]; k++) turn[0][j].Q[turn[0][j].cnt++] = k - 1;
    }
    int cur = 0;
    for (j = i; j < n; j++) {
      int flag = down((j - i) & 1, str[j]);
      if (pre[cur][str[j] - 'a'] == 0) {
        if (flag == 1) ans++;
        pre[cur][str[j] - 'a'] = allc++;
      }
      cur = pre[cur][str[j] - 'a'];
    }
  }
  printf("%d\n", ans);
  return 0;
}

```