#include <bits/stdc++.h>
using namespace std;
struct SA {
  struct state {
    int len, link, nxt[1 << 8], x = 0;
    state() : len(0), link(-1) { fill(begin(nxt), end(nxt), 0); }
  };
  vector<state> d;
  int last = 0;
  SA() { d.push_back(state()); }
  void add(char ch, int y) {
    int cur = d.size();
    d.push_back(state());
    d[cur].len = d[last].len + 1;
    d[cur].x = y;
    int p = last;
    while (p != -1 && !d[p].nxt[ch]) {
      d[p].nxt[ch] = cur;
      p = d[p].link;
    }
    if (p == -1) {
      d[cur].link = 0;
    } else {
      int q = d[p].nxt[ch];
      if (d[q].len == d[p].len + 1) {
        d[cur].link = q;
      } else {
        int clone = d.size();
        d.push_back(d[q]);
        d[clone].len = d[p].len + 1;
        d[clone].x = 11;
        while (p != -1 && d[p].nxt[ch] == q) {
          d[p].nxt[ch] = clone;
          p = d[p].link;
        }
        d[cur].link = d[q].link = clone;
      }
    }
    last = cur;
  }
} sa;
const int N = 12e5;
vector<int> adj[N];
int flg[N], n, l[12], r[12], cnt[N][12];
long long dp[N];
void dfs(int v) {
  cnt[v][sa.d[v].x]++;
  for (int u : adj[v]) {
    dfs(u);
    for (int j = 0; j <= n; j++) {
      cnt[v][j] += cnt[u][j];
    }
  }
  flg[v] = true;
  for (int i = 1; i <= n; i++) {
    flg[v] &= (l[i] <= cnt[v][i] && cnt[v][i] <= r[i]);
  }
}
long long f(int v) {
  long long ans = (cnt[v][0] > 0 && flg[v]) && v != 0;
  if (~dp[v]) {
    return dp[v];
  }
  for (int i = 'a'; i <= 'z'; i++)
    if (sa.d[v].nxt[i]) {
      ans += f(sa.d[v].nxt[i]);
    }
  return dp[v] = ans;
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string str;
  cin >> str;
  for (auto ch : str) {
    sa.add(ch, 0);
  }
  cin >> n;
  for (int i = 1; i <= n; i++) {
    sa.add('A' + i - 1, n + 1);
    cin >> str >> l[i] >> r[i];
    for (auto ch : str) {
      sa.add(ch, i);
    }
  }
  for (int i = 1; i < sa.d.size(); i++) {
    adj[sa.d[i].link].push_back(i);
  }
  dfs(0);
  memset(dp, -1, sizeof dp);
  cout << f(0) << endl;
}

### Prompt

Create a solution in CPP for 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 SA {
  struct state {
    int len, link, nxt[1 << 8], x = 0;
    state() : len(0), link(-1) { fill(begin(nxt), end(nxt), 0); }
  };
  vector<state> d;
  int last = 0;
  SA() { d.push_back(state()); }
  void add(char ch, int y) {
    int cur = d.size();
    d.push_back(state());
    d[cur].len = d[last].len + 1;
    d[cur].x = y;
    int p = last;
    while (p != -1 && !d[p].nxt[ch]) {
      d[p].nxt[ch] = cur;
      p = d[p].link;
    }
    if (p == -1) {
      d[cur].link = 0;
    } else {
      int q = d[p].nxt[ch];
      if (d[q].len == d[p].len + 1) {
        d[cur].link = q;
      } else {
        int clone = d.size();
        d.push_back(d[q]);
        d[clone].len = d[p].len + 1;
        d[clone].x = 11;
        while (p != -1 && d[p].nxt[ch] == q) {
          d[p].nxt[ch] = clone;
          p = d[p].link;
        }
        d[cur].link = d[q].link = clone;
      }
    }
    last = cur;
  }
} sa;
const int N = 12e5;
vector<int> adj[N];
int flg[N], n, l[12], r[12], cnt[N][12];
long long dp[N];
void dfs(int v) {
  cnt[v][sa.d[v].x]++;
  for (int u : adj[v]) {
    dfs(u);
    for (int j = 0; j <= n; j++) {
      cnt[v][j] += cnt[u][j];
    }
  }
  flg[v] = true;
  for (int i = 1; i <= n; i++) {
    flg[v] &= (l[i] <= cnt[v][i] && cnt[v][i] <= r[i]);
  }
}
long long f(int v) {
  long long ans = (cnt[v][0] > 0 && flg[v]) && v != 0;
  if (~dp[v]) {
    return dp[v];
  }
  for (int i = 'a'; i <= 'z'; i++)
    if (sa.d[v].nxt[i]) {
      ans += f(sa.d[v].nxt[i]);
    }
  return dp[v] = ans;
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string str;
  cin >> str;
  for (auto ch : str) {
    sa.add(ch, 0);
  }
  cin >> n;
  for (int i = 1; i <= n; i++) {
    sa.add('A' + i - 1, n + 1);
    cin >> str >> l[i] >> r[i];
    for (auto ch : str) {
      sa.add(ch, i);
    }
  }
  for (int i = 1; i < sa.d.size(); i++) {
    adj[sa.d[i].link].push_back(i);
  }
  dfs(0);
  memset(dp, -1, sizeof dp);
  cout << f(0) << endl;
}

```