#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int maxn = 2e3 + 10, maxm = 1e3 + 10;
const long long mod = 1e9 + 12341;
int n, m;
const long long decm = 128;
long long pw[maxn];
void init(int n = maxn - 5) {
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * decm % mod;
}
class strhash {
 public:
  long long hs[maxn], len;
  long long calhs(string &s, int n) {
    len = n;
    for (int i = 0; i <= len - 1; ++i) hs[i + 1] = (hs[i] * decm + s[i]) % mod;
    return hs[len];
  }
  inline long long geths(int a, int b) {
    return (hs[b] - (hs[a - 1] * pw[b - a + 1]) % mod + mod) % mod;
  }
} mps, mpt[20];
int l[20], r[20];
vector<long long> h;
unordered_map<long long, int> cnt[20];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init(maxn - 5);
  string s;
  cin >> s;
  cin >> n;
  mps.calhs(s, s.size());
  m = s.size();
  for (int i = 1; i <= m; ++i)
    for (int j = i; j <= m; ++j) h.push_back(mps.geths(i, j));
  sort(h.begin(), h.end());
  h.erase(unique(h.begin(), h.end()), h.end());
  for (int i = 1; i <= n; ++i) {
    string t;
    cin >> t >> l[i] >> r[i];
    mpt[i].calhs(t, t.size());
    int len = t.size();
    for (int j = 1; j <= len; ++j)
      for (int k = j; k <= len; ++k) cnt[i][mpt[i].geths(j, k)]++;
  }
  long long ans = 0;
  for (auto hs : h) {
    int flag = 0;
    for (int i = 1; i <= n; ++i) {
      int num = cnt[i][hs];
      if (num >= l[i] && num <= r[i]) flag++;
    }
    if (flag == n) ans++;
  }
  cout << ans << '\n';
}

### 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>
#pragma GCC optimize("Ofast")
using namespace std;
const int maxn = 2e3 + 10, maxm = 1e3 + 10;
const long long mod = 1e9 + 12341;
int n, m;
const long long decm = 128;
long long pw[maxn];
void init(int n = maxn - 5) {
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * decm % mod;
}
class strhash {
 public:
  long long hs[maxn], len;
  long long calhs(string &s, int n) {
    len = n;
    for (int i = 0; i <= len - 1; ++i) hs[i + 1] = (hs[i] * decm + s[i]) % mod;
    return hs[len];
  }
  inline long long geths(int a, int b) {
    return (hs[b] - (hs[a - 1] * pw[b - a + 1]) % mod + mod) % mod;
  }
} mps, mpt[20];
int l[20], r[20];
vector<long long> h;
unordered_map<long long, int> cnt[20];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init(maxn - 5);
  string s;
  cin >> s;
  cin >> n;
  mps.calhs(s, s.size());
  m = s.size();
  for (int i = 1; i <= m; ++i)
    for (int j = i; j <= m; ++j) h.push_back(mps.geths(i, j));
  sort(h.begin(), h.end());
  h.erase(unique(h.begin(), h.end()), h.end());
  for (int i = 1; i <= n; ++i) {
    string t;
    cin >> t >> l[i] >> r[i];
    mpt[i].calhs(t, t.size());
    int len = t.size();
    for (int j = 1; j <= len; ++j)
      for (int k = j; k <= len; ++k) cnt[i][mpt[i].geths(j, k)]++;
  }
  long long ans = 0;
  for (auto hs : h) {
    int flag = 0;
    for (int i = 1; i <= n; ++i) {
      int num = cnt[i][hs];
      if (num >= l[i] && num <= r[i]) flag++;
    }
    if (flag == n) ans++;
  }
  cout << ans << '\n';
}

```