#include <bits/stdc++.h>
using namespace std;
const int N = 1110000;
int ch[N][26], prt[N], len[N], Right[N][11];
int L[21], R[21], n;
int Lst, root, node;
char s[12][51000];
int newnode(int l) {
  int x = ++node;
  len[x] = l;
  return x;
}
void Insert(int w) {
  int p = Lst, nx = newnode(len[p] + 1);
  Lst = nx;
  for (; p && !ch[p][w]; p = prt[p]) ch[p][w] = nx;
  if (!p)
    prt[nx] = root;
  else {
    int q = ch[p][w];
    if (len[q] == len[p] + 1)
      prt[nx] = q;
    else {
      int newq = newnode(len[p] + 1);
      for (int i = 0; i < 26; ++i) ch[newq][i] = ch[q][i];
      prt[newq] = prt[q];
      prt[nx] = prt[q] = newq;
      for (; p && ch[p][w] == q; p = prt[p]) ch[p][w] = newq;
    }
  }
}
vector<int> G[N];
void DFS(int u) {
  for (int i = 0; i < G[u].size(); ++i) {
    int to = G[u][i];
    DFS(to);
    for (int k = 0; k <= n; ++k) Right[u][k] += Right[to][k];
  }
}
int main() {
  scanf("%s%d", s[0], &n);
  for (int i = 1; i <= n; ++i) scanf("%s%d%d", &s[i], &L[i], &R[i]);
  root = Lst = newnode(0);
  for (int i = 0; i <= n; ++i) {
    Lst = root;
    int l = strlen(s[i]);
    for (int j = 0; j < l; ++j) {
      Insert(s[i][j] - 'a');
      Right[Lst][i]++;
    }
  }
  for (int i = 2; i <= node; ++i) G[prt[i]].push_back(i);
  DFS(root);
  long long Ans = 0;
  for (int i = 2; i <= node; ++i) {
    if (!Right[i][0]) continue;
    bool flag = 1;
    for (int k = 1; k <= n; ++k)
      if (Right[i][k] < L[k] || Right[i][k] > R[k]) {
        flag = 0;
        break;
      }
    if (flag) Ans += 1ll * (len[i] - len[prt[i]]);
  }
  cout << Ans << '\n';
  return 0;
}

### Prompt

Please formulate 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 = 1110000;
int ch[N][26], prt[N], len[N], Right[N][11];
int L[21], R[21], n;
int Lst, root, node;
char s[12][51000];
int newnode(int l) {
  int x = ++node;
  len[x] = l;
  return x;
}
void Insert(int w) {
  int p = Lst, nx = newnode(len[p] + 1);
  Lst = nx;
  for (; p && !ch[p][w]; p = prt[p]) ch[p][w] = nx;
  if (!p)
    prt[nx] = root;
  else {
    int q = ch[p][w];
    if (len[q] == len[p] + 1)
      prt[nx] = q;
    else {
      int newq = newnode(len[p] + 1);
      for (int i = 0; i < 26; ++i) ch[newq][i] = ch[q][i];
      prt[newq] = prt[q];
      prt[nx] = prt[q] = newq;
      for (; p && ch[p][w] == q; p = prt[p]) ch[p][w] = newq;
    }
  }
}
vector<int> G[N];
void DFS(int u) {
  for (int i = 0; i < G[u].size(); ++i) {
    int to = G[u][i];
    DFS(to);
    for (int k = 0; k <= n; ++k) Right[u][k] += Right[to][k];
  }
}
int main() {
  scanf("%s%d", s[0], &n);
  for (int i = 1; i <= n; ++i) scanf("%s%d%d", &s[i], &L[i], &R[i]);
  root = Lst = newnode(0);
  for (int i = 0; i <= n; ++i) {
    Lst = root;
    int l = strlen(s[i]);
    for (int j = 0; j < l; ++j) {
      Insert(s[i][j] - 'a');
      Right[Lst][i]++;
    }
  }
  for (int i = 2; i <= node; ++i) G[prt[i]].push_back(i);
  DFS(root);
  long long Ans = 0;
  for (int i = 2; i <= node; ++i) {
    if (!Right[i][0]) continue;
    bool flag = 1;
    for (int k = 1; k <= n; ++k)
      if (Right[i][k] < L[k] || Right[i][k] > R[k]) {
        flag = 0;
        break;
      }
    if (flag) Ans += 1ll * (len[i] - len[prt[i]]);
  }
  cout << Ans << '\n';
  return 0;
}

```