#include <bits/stdc++.h>
using namespace std;
using ll = long long;
inline int read();
const int M = 500016, MOD = 1000000007;
int sz, lst;
int ch[M << 1][40], len[M << 1], link[M << 1], cnt[12][M << 1];
void extend(const char *s, int id) {
  for (int i = 0; s[i]; ++i) {
    int c = s[i] - 'a';
    int cur = ++sz;
    len[cur] = len[lst] + 1;
    int p = lst;
    while (!ch[p][c]) {
      ch[p][c] = cur;
      p = link[p];
    }
    if (ch[p][c] != cur) {
      int q = ch[p][c];
      if (len[p] + 1 == len[q])
        link[cur] = q;
      else {
        int clone = ++sz;
        memcpy(ch[clone], ch[q], sizeof(ch[q]));
        link[clone] = link[q];
        len[clone] = len[p] + 1;
        while (ch[p][c] == q) {
          ch[p][c] = clone;
          p = link[p];
        }
        link[q] = link[cur] = clone;
      }
    }
    lst = cur;
    ++cnt[id][cur];
  }
}
void build() {
  vector<int> nodes;
  for (int i = 1; i <= sz; ++i) nodes.push_back(i);
  sort(nodes.begin(), nodes.end(),
       [&](int a, int b) { return len[a] > len[b]; });
  for (auto u : nodes)
    for (int i = 0; i < 12; ++i) cnt[i][link[u]] += cnt[i][u];
}
char str[M];
int lef[M], rig[M];
int main(void) {
  scanf("%s", str);
  extend(str, 0);
  lef[0] = 1, rig[0] = 50000;
  int n = read();
  for (int i = 1; i <= n; ++i) {
    string t = "z";
    t[0] += i;
    extend(t.c_str(), n + 1);
    scanf("%s", str);
    extend(str, i);
    lef[i] = read(), rig[i] = read();
  }
  build();
  int ans = 0;
  for (int i = 1; i <= sz; ++i) {
    int suc = 1;
    for (int j = 0; j <= n; ++j)
      if (cnt[j][i] < lef[j] || cnt[j][i] > rig[j]) suc = 0, j = n + 1;
    ans += suc * (len[i] - len[link[i]]);
  }
  printf("%d\n", ans);
  return 0;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}

### Prompt

Please provide a CPP coded solution 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;
using ll = long long;
inline int read();
const int M = 500016, MOD = 1000000007;
int sz, lst;
int ch[M << 1][40], len[M << 1], link[M << 1], cnt[12][M << 1];
void extend(const char *s, int id) {
  for (int i = 0; s[i]; ++i) {
    int c = s[i] - 'a';
    int cur = ++sz;
    len[cur] = len[lst] + 1;
    int p = lst;
    while (!ch[p][c]) {
      ch[p][c] = cur;
      p = link[p];
    }
    if (ch[p][c] != cur) {
      int q = ch[p][c];
      if (len[p] + 1 == len[q])
        link[cur] = q;
      else {
        int clone = ++sz;
        memcpy(ch[clone], ch[q], sizeof(ch[q]));
        link[clone] = link[q];
        len[clone] = len[p] + 1;
        while (ch[p][c] == q) {
          ch[p][c] = clone;
          p = link[p];
        }
        link[q] = link[cur] = clone;
      }
    }
    lst = cur;
    ++cnt[id][cur];
  }
}
void build() {
  vector<int> nodes;
  for (int i = 1; i <= sz; ++i) nodes.push_back(i);
  sort(nodes.begin(), nodes.end(),
       [&](int a, int b) { return len[a] > len[b]; });
  for (auto u : nodes)
    for (int i = 0; i < 12; ++i) cnt[i][link[u]] += cnt[i][u];
}
char str[M];
int lef[M], rig[M];
int main(void) {
  scanf("%s", str);
  extend(str, 0);
  lef[0] = 1, rig[0] = 50000;
  int n = read();
  for (int i = 1; i <= n; ++i) {
    string t = "z";
    t[0] += i;
    extend(t.c_str(), n + 1);
    scanf("%s", str);
    extend(str, i);
    lef[i] = read(), rig[i] = read();
  }
  build();
  int ans = 0;
  for (int i = 1; i <= sz; ++i) {
    int suc = 1;
    for (int j = 0; j <= n; ++j)
      if (cnt[j][i] < lef[j] || cnt[j][i] > rig[j]) suc = 0, j = n + 1;
    ans += suc * (len[i] - len[link[i]]);
  }
  printf("%d\n", ans);
  return 0;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}

```