#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
struct HASH {
  long long H1, H2;
  HASH(long long H1 = 0, long long H2 = 0) {
    this->H1 = H1;
    this->H2 = H2;
  }
};
char Str[2007 + 7];
long long Len;
long long N;
long long HS1[2007 + 7];
long long HS2[2007 + 7];
long long Pow1[2007 + 7];
long long Pow2[2007 + 7];
long long Inv1[2007 + 7];
long long Inv2[2007 + 7];
long long BASE1 = 31;
long long BASE2 = 47;
long long MOD1 = 1000000007;
long long MOD2 = 1000000007;
vector<HASH> Subs[2007 + 7];
vector<bool> NtP[2007 + 7];
map<HASH, long long> Cnt[2007 + 7];
bool operator<(const HASH &a, const HASH &b) {
  if (a.H1 != b.H1)
    return a.H1 < b.H1;
  else
    return a.H2 < b.H2;
}
bool operator==(const HASH &a, const HASH &b) {
  return a.H1 == b.H1 && a.H2 == b.H2;
}
void SetHash(void) {
  long long i;
  for (i = 1; i <= Len; i++) {
    HS1[i] = (Str[i] * Pow1[i]) % MOD1;
    HS2[i] = (Str[i] * Pow2[i]) % MOD2;
  }
  for (i = 1; i <= Len; i++) {
    HS1[i] = (HS1[i] + HS1[i - 1]) % MOD1;
    HS2[i] = (HS2[i] + HS2[i - 1]) % MOD2;
  }
}
long long Pow(long long a, long long p, long long m) {
  if (!p)
    return 1;
  else if (p & 1) {
    return (Pow(a, p - 1, m) * a) % m;
  } else {
    long long v = Pow(a, p / 2, m);
    return (v * v) % m;
  }
}
int main(void) {
  long long i, j, v, l, h, H1, H2, Icase, k = 0;
  Pow1[0] = Inv1[0] = 1;
  Pow2[0] = Inv2[0] = 1;
  for (i = 1; i <= 2007; i++) {
    Pow1[i] = (BASE1 * Pow1[i - 1]) % MOD1;
    Inv1[i] = Pow(Pow1[i], MOD1 - 2, MOD1);
    Pow2[i] = (BASE2 * Pow2[i - 1]) % MOD2;
    Inv2[i] = Pow(Pow2[i], MOD2 - 2, MOD2);
  }
  scanf("%s", &Str[1]);
  Len = strlen(&Str[1]);
  long long InitL = Len;
  SetHash();
  for (i = 1; i <= Len; i++) {
    for (j = i; j <= Len; j++) {
      H1 = (HS1[j] - HS1[i - 1] + MOD1) % MOD1;
      H1 = (H1 * Inv1[i]) % MOD1;
      H2 = (HS2[j] - HS2[i - 1] + MOD2) % MOD2;
      H2 = (H2 * Inv2[i]) % MOD2;
      Subs[j - i].push_back(HASH(H1, H2));
    }
  }
  for (i = 0; i <= Len; i++) {
    sort(Subs[i].begin(), Subs[i].end());
    Subs[i].resize(unique(Subs[i].begin(), Subs[i].end()) - Subs[i].begin());
    NtP[i].resize(Subs[i].size());
  }
  scanf("%I64d", &N);
  for (k = 1; k <= N; k++) {
    scanf("%s%I64d%I64d", &Str[1], &l, &h);
    Len = strlen(&Str[1]);
    SetHash();
    for (i = 1; i <= Len; i++) {
      for (j = i; j <= Len; j++) {
        H1 = (HS1[j] - HS1[i - 1] + MOD1) % MOD1;
        H1 = (H1 * Inv1[i]) % MOD1;
        H2 = (HS2[j] - HS2[i - 1] + MOD2) % MOD2;
        H2 = (H2 * Inv2[i]) % MOD2;
        Cnt[j - i][HASH(H1, H2)]++;
      }
    }
    for (i = 0; i <= InitL; i++) {
      for (j = 0; j < Subs[i].size(); j++) {
        if (NtP[i][j]) continue;
        if (Cnt[i].find(Subs[i][j]) == Cnt[i].end())
          v = 0;
        else
          v = Cnt[i][Subs[i][j]];
        if (v < l || v > h) {
          NtP[i][j] = true;
          continue;
        }
      }
      Cnt[i].clear();
    }
  }
  long long Ans = 0;
  for (i = 0; i <= InitL; i++) {
    for (j = 0; j < Subs[i].size(); j++) {
      if (NtP[i][j]) continue;
      Ans++;
    }
  }
  printf("%I64d\n", Ans);
  return 0;
}

### Prompt

Generate 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 comment(linker, "/STACK:16777216")
using namespace std;
struct HASH {
  long long H1, H2;
  HASH(long long H1 = 0, long long H2 = 0) {
    this->H1 = H1;
    this->H2 = H2;
  }
};
char Str[2007 + 7];
long long Len;
long long N;
long long HS1[2007 + 7];
long long HS2[2007 + 7];
long long Pow1[2007 + 7];
long long Pow2[2007 + 7];
long long Inv1[2007 + 7];
long long Inv2[2007 + 7];
long long BASE1 = 31;
long long BASE2 = 47;
long long MOD1 = 1000000007;
long long MOD2 = 1000000007;
vector<HASH> Subs[2007 + 7];
vector<bool> NtP[2007 + 7];
map<HASH, long long> Cnt[2007 + 7];
bool operator<(const HASH &a, const HASH &b) {
  if (a.H1 != b.H1)
    return a.H1 < b.H1;
  else
    return a.H2 < b.H2;
}
bool operator==(const HASH &a, const HASH &b) {
  return a.H1 == b.H1 && a.H2 == b.H2;
}
void SetHash(void) {
  long long i;
  for (i = 1; i <= Len; i++) {
    HS1[i] = (Str[i] * Pow1[i]) % MOD1;
    HS2[i] = (Str[i] * Pow2[i]) % MOD2;
  }
  for (i = 1; i <= Len; i++) {
    HS1[i] = (HS1[i] + HS1[i - 1]) % MOD1;
    HS2[i] = (HS2[i] + HS2[i - 1]) % MOD2;
  }
}
long long Pow(long long a, long long p, long long m) {
  if (!p)
    return 1;
  else if (p & 1) {
    return (Pow(a, p - 1, m) * a) % m;
  } else {
    long long v = Pow(a, p / 2, m);
    return (v * v) % m;
  }
}
int main(void) {
  long long i, j, v, l, h, H1, H2, Icase, k = 0;
  Pow1[0] = Inv1[0] = 1;
  Pow2[0] = Inv2[0] = 1;
  for (i = 1; i <= 2007; i++) {
    Pow1[i] = (BASE1 * Pow1[i - 1]) % MOD1;
    Inv1[i] = Pow(Pow1[i], MOD1 - 2, MOD1);
    Pow2[i] = (BASE2 * Pow2[i - 1]) % MOD2;
    Inv2[i] = Pow(Pow2[i], MOD2 - 2, MOD2);
  }
  scanf("%s", &Str[1]);
  Len = strlen(&Str[1]);
  long long InitL = Len;
  SetHash();
  for (i = 1; i <= Len; i++) {
    for (j = i; j <= Len; j++) {
      H1 = (HS1[j] - HS1[i - 1] + MOD1) % MOD1;
      H1 = (H1 * Inv1[i]) % MOD1;
      H2 = (HS2[j] - HS2[i - 1] + MOD2) % MOD2;
      H2 = (H2 * Inv2[i]) % MOD2;
      Subs[j - i].push_back(HASH(H1, H2));
    }
  }
  for (i = 0; i <= Len; i++) {
    sort(Subs[i].begin(), Subs[i].end());
    Subs[i].resize(unique(Subs[i].begin(), Subs[i].end()) - Subs[i].begin());
    NtP[i].resize(Subs[i].size());
  }
  scanf("%I64d", &N);
  for (k = 1; k <= N; k++) {
    scanf("%s%I64d%I64d", &Str[1], &l, &h);
    Len = strlen(&Str[1]);
    SetHash();
    for (i = 1; i <= Len; i++) {
      for (j = i; j <= Len; j++) {
        H1 = (HS1[j] - HS1[i - 1] + MOD1) % MOD1;
        H1 = (H1 * Inv1[i]) % MOD1;
        H2 = (HS2[j] - HS2[i - 1] + MOD2) % MOD2;
        H2 = (H2 * Inv2[i]) % MOD2;
        Cnt[j - i][HASH(H1, H2)]++;
      }
    }
    for (i = 0; i <= InitL; i++) {
      for (j = 0; j < Subs[i].size(); j++) {
        if (NtP[i][j]) continue;
        if (Cnt[i].find(Subs[i][j]) == Cnt[i].end())
          v = 0;
        else
          v = Cnt[i][Subs[i][j]];
        if (v < l || v > h) {
          NtP[i][j] = true;
          continue;
        }
      }
      Cnt[i].clear();
    }
  }
  long long Ans = 0;
  for (i = 0; i <= InitL; i++) {
    for (j = 0; j < Subs[i].size(); j++) {
      if (NtP[i][j]) continue;
      Ans++;
    }
  }
  printf("%I64d\n", Ans);
  return 0;
}

```