#include <bits/stdc++.h>
using namespace std;
const int N = 50004;
const int V = N * 2;
const int A = 26;
int ololink[V], len[V], to[V][26], d;
int cnt[V];
vector<int> v[N];
int sl[N], sr[N];
void sau(char *s, int n, bool fl) {
  d = 0;
  int l = d++;
  memset(to[l], 0, sizeof(to[l]));
  ololink[l] = -1;
  cnt[l] = 0;
  for (int i = 0; i < n; ++i) {
    int c = s[i] - 'a';
    int x = d++;
    memset(to[x], 0, sizeof(to[x]));
    len[x] = len[l] + 1;
    ololink[x] = 0;
    cnt[x] = 1;
    for (; l != -1 && !to[l][c]; l = ololink[l]) to[l][c] = x;
    if (l != -1) {
      int p = to[l][c];
      if (len[p] == len[l] + 1)
        ololink[x] = p;
      else {
        int q = d++;
        len[q] = len[l] + 1;
        ololink[q] = ololink[p];
        memcpy(to[q], to[p], sizeof(to[q]));
        cnt[q] = 0;
        for (; l != -1 && to[l][c] == p; l = ololink[l]) to[l][c] = q;
        ololink[x] = ololink[p] = q;
      }
    }
    l = x;
    if (fl) {
      sl[i] = 0;
      sr[i] = i - len[ololink[l]];
    }
  }
}
char s[N], p[N];
void docnt(char *s, int n) {
  for (int i = 0; i <= n; ++i) v[i].clear();
  for (int i = 0; i < d; ++i) v[len[i]].push_back(i);
  for (int l = n; l; --l)
    for (int k = 0; k < v[l].size(); ++k) {
      int i = v[l][k];
      int j = ololink[i];
      cnt[j] += cnt[i];
    }
}
int main() {
  int n, m, q;
  scanf("%s", &s);
  n = strlen(s);
  sau(s, n, true);
  scanf("%d", &q);
  for (int k = 0; k < q; ++k) {
    int L, R;
    scanf("%s%d%d", &p, &L, &R);
    m = strlen(p);
    sau(p, m, false);
    docnt(p, m);
    int vl, vr;
    int dl, dr;
    vl = vr = 0;
    dl = dr = 0;
    for (int i = 0; i < n; ++i) {
      int c = s[i] - 'a';
      while (vl != -1 && !to[vl][c]) vl = ololink[vl], dl = min(dl, len[vl]);
      if (vl != -1) {
        vl = to[vl][c];
        ++dl;
        while (vl != -1 && cnt[vl] < L) vl = ololink[vl], dl = min(dl, len[vl]);
        if (vl == -1) dl = vl = 0;
      } else {
        vl = 0;
        dl = 0;
      }
      while (vr != -1 && !to[vr][c]) vr = ololink[vr], dr = min(dr, len[vr]);
      if (vr != -1) {
        vr = to[vr][c];
        ++dr;
        while (vr != -1 && cnt[vr] < R + 1)
          vr = ololink[vr], dr = min(dr, len[vr]);
        if (vr == -1) dr = vr = 0;
      } else {
        vr = 0;
        dr = 0;
      }
      int kl = dl, kr = dr;
      int cl = cnt[vl], cr = cnt[vr];
      if (L == 0) {
        if (kl < i + 1) {
          kl = i + 1;
          cr = 0;
        }
      }
      ++kr;
      sl[i] = max(sl[i], i - kl + 1);
      sr[i] = min(sr[i], i - kr + 1);
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans += max(0, sr[i] - sl[i] + 1);
  cout << ans << endl;
  return 0;
}

### 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;
const int N = 50004;
const int V = N * 2;
const int A = 26;
int ololink[V], len[V], to[V][26], d;
int cnt[V];
vector<int> v[N];
int sl[N], sr[N];
void sau(char *s, int n, bool fl) {
  d = 0;
  int l = d++;
  memset(to[l], 0, sizeof(to[l]));
  ololink[l] = -1;
  cnt[l] = 0;
  for (int i = 0; i < n; ++i) {
    int c = s[i] - 'a';
    int x = d++;
    memset(to[x], 0, sizeof(to[x]));
    len[x] = len[l] + 1;
    ololink[x] = 0;
    cnt[x] = 1;
    for (; l != -1 && !to[l][c]; l = ololink[l]) to[l][c] = x;
    if (l != -1) {
      int p = to[l][c];
      if (len[p] == len[l] + 1)
        ololink[x] = p;
      else {
        int q = d++;
        len[q] = len[l] + 1;
        ololink[q] = ololink[p];
        memcpy(to[q], to[p], sizeof(to[q]));
        cnt[q] = 0;
        for (; l != -1 && to[l][c] == p; l = ololink[l]) to[l][c] = q;
        ololink[x] = ololink[p] = q;
      }
    }
    l = x;
    if (fl) {
      sl[i] = 0;
      sr[i] = i - len[ololink[l]];
    }
  }
}
char s[N], p[N];
void docnt(char *s, int n) {
  for (int i = 0; i <= n; ++i) v[i].clear();
  for (int i = 0; i < d; ++i) v[len[i]].push_back(i);
  for (int l = n; l; --l)
    for (int k = 0; k < v[l].size(); ++k) {
      int i = v[l][k];
      int j = ololink[i];
      cnt[j] += cnt[i];
    }
}
int main() {
  int n, m, q;
  scanf("%s", &s);
  n = strlen(s);
  sau(s, n, true);
  scanf("%d", &q);
  for (int k = 0; k < q; ++k) {
    int L, R;
    scanf("%s%d%d", &p, &L, &R);
    m = strlen(p);
    sau(p, m, false);
    docnt(p, m);
    int vl, vr;
    int dl, dr;
    vl = vr = 0;
    dl = dr = 0;
    for (int i = 0; i < n; ++i) {
      int c = s[i] - 'a';
      while (vl != -1 && !to[vl][c]) vl = ololink[vl], dl = min(dl, len[vl]);
      if (vl != -1) {
        vl = to[vl][c];
        ++dl;
        while (vl != -1 && cnt[vl] < L) vl = ololink[vl], dl = min(dl, len[vl]);
        if (vl == -1) dl = vl = 0;
      } else {
        vl = 0;
        dl = 0;
      }
      while (vr != -1 && !to[vr][c]) vr = ololink[vr], dr = min(dr, len[vr]);
      if (vr != -1) {
        vr = to[vr][c];
        ++dr;
        while (vr != -1 && cnt[vr] < R + 1)
          vr = ololink[vr], dr = min(dr, len[vr]);
        if (vr == -1) dr = vr = 0;
      } else {
        vr = 0;
        dr = 0;
      }
      int kl = dl, kr = dr;
      int cl = cnt[vl], cr = cnt[vr];
      if (L == 0) {
        if (kl < i + 1) {
          kl = i + 1;
          cr = 0;
        }
      }
      ++kr;
      sl[i] = max(sl[i], i - kl + 1);
      sr[i] = min(sr[i], i - kr + 1);
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans += max(0, sr[i] - sl[i] + 1);
  cout << ans << endl;
  return 0;
}

```