#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5;
const int maxk = 20;
const long long P = 1e9 + 7;
int n, k;
char s[maxn];
int l[maxn], r[maxn];
char p[maxk][maxn];
int lenp[maxk];
long long lkdfjldskfj[maxn];
long long hashp[maxk][maxn];
long long powers[maxn];
vector<int> pos[maxk], npos[maxk];
inline long long gethash(int l, int r) {
  return lkdfjldskfj[r] - lkdfjldskfj[l - 1] * powers[r - l + 1];
}
inline long long gethashp(int l, int r, int num) {
  return hashp[num][r] - hashp[num][l - 1] * powers[r - l + 1];
}
inline bool eq(int l, int r, int lp, int rp, int it) {
  return gethash(l + 1, r + 1) == gethashp(lp + 1, rp + 1, it);
}
int main() {
  powers[0] = 1;
  for (int i = 1; i < maxn; i++) powers[i] = powers[i - 1] * P;
  gets(s);
  n = strlen(s);
  lkdfjldskfj[0] = 0;
  for (int i = 0; i < n; i++) lkdfjldskfj[i + 1] = lkdfjldskfj[i] * P + s[i];
  scanf("%d\n", &k);
  for (int i = 0; i < k; i++) {
    scanf("%s%d%d", p[i], &l[i], &r[i]), lenp[i] = strlen(p[i]);
    hashp[i][0] = 0;
    for (int j = 0; j < lenp[i]; j++)
      hashp[i][j + 1] = hashp[i][j] * P + p[i][j];
  }
  set<long long> ans;
  for (int lp = 0; lp < n; lp++) {
    bool fail = 0;
    for (int i = 0; i < k; i++) {
      pos[i].clear();
      pos[i].resize(lenp[i]);
      for (int j = 0; j < lenp[i]; j++) pos[i][j] = j;
    }
    for (int rp = lp; rp < n && !fail; rp++) {
      long long h = gethash(lp + 1, rp + 1);
      if (ans.find(h) != ans.end()) continue;
      int len = rp - lp + 1;
      bool ok = 1;
      for (int it = 0; it < k; it++) {
        npos[it].clear();
        int cnt = 0;
        for (int i = 0; i < (int)pos[it].size(); i++)
          if (pos[it][i] + len <= lenp[it] &&
              eq(lp, rp, pos[it][i], pos[it][i] + len - 1, it)) {
            ++cnt, npos[it].push_back(pos[it][i]);
            if (cnt > r[it]) break;
          }
        if (cnt < l[it]) {
          ok = 0;
          fail = 1;
          break;
        }
        if (cnt > r[it]) {
          ok = 0;
          break;
        }
        pos[it].resize((int)npos[it].size());
        for (int i = 0; i < (int)npos[it].size(); i++) pos[it][i] = npos[it][i];
      }
      if (ok) {
        ans.insert(h);
      }
    }
  }
  printf("%d\n", (int)ans.size());
  return 0;
}

### Prompt

Your challenge is to write 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 maxn = 2e5;
const int maxk = 20;
const long long P = 1e9 + 7;
int n, k;
char s[maxn];
int l[maxn], r[maxn];
char p[maxk][maxn];
int lenp[maxk];
long long lkdfjldskfj[maxn];
long long hashp[maxk][maxn];
long long powers[maxn];
vector<int> pos[maxk], npos[maxk];
inline long long gethash(int l, int r) {
  return lkdfjldskfj[r] - lkdfjldskfj[l - 1] * powers[r - l + 1];
}
inline long long gethashp(int l, int r, int num) {
  return hashp[num][r] - hashp[num][l - 1] * powers[r - l + 1];
}
inline bool eq(int l, int r, int lp, int rp, int it) {
  return gethash(l + 1, r + 1) == gethashp(lp + 1, rp + 1, it);
}
int main() {
  powers[0] = 1;
  for (int i = 1; i < maxn; i++) powers[i] = powers[i - 1] * P;
  gets(s);
  n = strlen(s);
  lkdfjldskfj[0] = 0;
  for (int i = 0; i < n; i++) lkdfjldskfj[i + 1] = lkdfjldskfj[i] * P + s[i];
  scanf("%d\n", &k);
  for (int i = 0; i < k; i++) {
    scanf("%s%d%d", p[i], &l[i], &r[i]), lenp[i] = strlen(p[i]);
    hashp[i][0] = 0;
    for (int j = 0; j < lenp[i]; j++)
      hashp[i][j + 1] = hashp[i][j] * P + p[i][j];
  }
  set<long long> ans;
  for (int lp = 0; lp < n; lp++) {
    bool fail = 0;
    for (int i = 0; i < k; i++) {
      pos[i].clear();
      pos[i].resize(lenp[i]);
      for (int j = 0; j < lenp[i]; j++) pos[i][j] = j;
    }
    for (int rp = lp; rp < n && !fail; rp++) {
      long long h = gethash(lp + 1, rp + 1);
      if (ans.find(h) != ans.end()) continue;
      int len = rp - lp + 1;
      bool ok = 1;
      for (int it = 0; it < k; it++) {
        npos[it].clear();
        int cnt = 0;
        for (int i = 0; i < (int)pos[it].size(); i++)
          if (pos[it][i] + len <= lenp[it] &&
              eq(lp, rp, pos[it][i], pos[it][i] + len - 1, it)) {
            ++cnt, npos[it].push_back(pos[it][i]);
            if (cnt > r[it]) break;
          }
        if (cnt < l[it]) {
          ok = 0;
          fail = 1;
          break;
        }
        if (cnt > r[it]) {
          ok = 0;
          break;
        }
        pos[it].resize((int)npos[it].size());
        for (int i = 0; i < (int)npos[it].size(); i++) pos[it][i] = npos[it][i];
      }
      if (ok) {
        ans.insert(h);
      }
    }
  }
  printf("%d\n", (int)ans.size());
  return 0;
}

```