#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
const int logN = 19;
const int MAX = 1e9 + 5;
int bef_size, LOG[N], bef[N][11], sorted[N], l, suff[N], rmq[N][logN + 1], k, n,
    q, dp[N][11], w[N], list[N], m, ll, ans, a[N][2];
string str, s, ss;
pair<pair<int, int>, int> C[N];
int query(int x, int y) {
  if (x == y) return n - sorted[x] + 1;
  y--;
  int t = LOG[y - x + 1];
  return min(rmq[x][t], rmq[y - (1 << t) + 1][t]);
}
bool ctr(int b, int l, bool t) {
  int bas = 1, son = b;
  while (bas < son) {
    int orta = bas + son >> 1;
    if (orta == bas) orta++;
    if (query(b - orta + 1, b) >= l)
      bas = orta;
    else
      son = orta - 1;
  }
  int top = b - bas + 1;
  bas = 1, son = n - b + 1;
  while (bas < son) {
    int orta = bas + son >> 1;
    if (bas == orta) orta++;
    if (query(b, b + orta - 1) >= l)
      bas = orta;
    else
      son = orta - 1;
  }
  int down = b + son - 1;
  if (t == false)
    for (int i = 1; i <= q; i++)
      if (dp[down][i] - dp[top - 1][i] < a[i][0]) return false;
  if (t == true)
    for (int i = 1; i <= q; i++)
      if (dp[down][i] - dp[top - 1][i] > a[i][1]) return false;
  return true;
}
int bs_left(int b) {
  int bas = 1, son = bef_size - sorted[b] + 1;
  while (bas < son) {
    int orta = bas + son >> 1;
    if (bas == orta) orta++;
    int tt = ctr(b, orta, 0);
    if (tt)
      bas = orta;
    else
      son = orta - 1;
  }
  return bas;
}
int bs_right(int b) {
  int bas = 1, son = bef_size - sorted[b] + 1;
  while (bas < son) {
    int orta = bas + son >> 1;
    int tt = ctr(b, orta, 1);
    if (!tt)
      bas = orta + 1;
    else
      son = orta;
  }
  return bas;
}
int main() {
  cin >> s;
  scanf("%d", &q);
  s = '0' + s;
  int ll = s.size();
  bef_size = s.size() - 1;
  for (int i = 1; i <= q; i++) {
    cin >> ss;
    scanf("%d %d", &a[i][0], &a[i][1]);
    int temp = ss.size();
    w[ll] = -1;
    for (int j = ll + 1; j <= ll + temp; j++) w[j] = i;
    s = s + '$' + ss;
    ll = s.size();
  }
  n = ll - 1;
  for (int i = 1; i <= n; i++) suff[i] = s[i];
  for (int i = 1; i <= n; i++) LOG[i] = log2(i);
  for (int j = 1; j <= logN; j++) {
    for (int i = 1; i <= n; i++)
      C[i] =
          make_pair(make_pair(suff[i], suff[min(i + (1 << j - 1), n + 1)]), i);
    sort(C + 1, C + n + 1);
    for (int i = 1; i <= n; i++)
      suff[C[i].second] =
          suff[C[i - 1].second] + (C[i - 1].first != C[i].first);
  }
  for (int i = 1; i <= n; i++) sorted[suff[i]] = i;
  int j = 0;
  for (int i = 1; i <= n; i++) {
    if (suff[i] == 1) continue;
    while (i + j <= n && sorted[suff[i] - 1] + j <= n &&
           s[sorted[suff[i] - 1] + j] == s[i + j])
      j++;
    rmq[suff[i] - 1][0] = j;
    if (j) j--;
  }
  rmq[n][0] = 1000000000;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= q; j++) {
      dp[i][j] = dp[i - 1][j] + (w[sorted[i]] == j);
      if (w[sorted[i]] == j)
        bef[i][j] = i;
      else
        bef[i][j] = bef[i - 1][j];
    }
  for (int j = 1; j <= logN; j++)
    for (int i = 1; i <= n; i++)
      rmq[i][j] = min(rmq[i][j - 1], rmq[min(n, i + (1 << j - 1))][j - 1]);
  int r = 0, l = 0, bef = 0;
  for (int i = 1; i <= n; i++) {
    if (w[sorted[i]] != 0) continue;
    l = bs_left(i);
    if (bef)
      r = max(bs_right(i), query(bef, i) + 1);
    else
      r = bs_right(i);
    bef = i;
    if (ctr(i, l, 0) && ctr(i, r, 1)) ans += max(l - r + 1, 0);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 = 5e5 + 5;
const int logN = 19;
const int MAX = 1e9 + 5;
int bef_size, LOG[N], bef[N][11], sorted[N], l, suff[N], rmq[N][logN + 1], k, n,
    q, dp[N][11], w[N], list[N], m, ll, ans, a[N][2];
string str, s, ss;
pair<pair<int, int>, int> C[N];
int query(int x, int y) {
  if (x == y) return n - sorted[x] + 1;
  y--;
  int t = LOG[y - x + 1];
  return min(rmq[x][t], rmq[y - (1 << t) + 1][t]);
}
bool ctr(int b, int l, bool t) {
  int bas = 1, son = b;
  while (bas < son) {
    int orta = bas + son >> 1;
    if (orta == bas) orta++;
    if (query(b - orta + 1, b) >= l)
      bas = orta;
    else
      son = orta - 1;
  }
  int top = b - bas + 1;
  bas = 1, son = n - b + 1;
  while (bas < son) {
    int orta = bas + son >> 1;
    if (bas == orta) orta++;
    if (query(b, b + orta - 1) >= l)
      bas = orta;
    else
      son = orta - 1;
  }
  int down = b + son - 1;
  if (t == false)
    for (int i = 1; i <= q; i++)
      if (dp[down][i] - dp[top - 1][i] < a[i][0]) return false;
  if (t == true)
    for (int i = 1; i <= q; i++)
      if (dp[down][i] - dp[top - 1][i] > a[i][1]) return false;
  return true;
}
int bs_left(int b) {
  int bas = 1, son = bef_size - sorted[b] + 1;
  while (bas < son) {
    int orta = bas + son >> 1;
    if (bas == orta) orta++;
    int tt = ctr(b, orta, 0);
    if (tt)
      bas = orta;
    else
      son = orta - 1;
  }
  return bas;
}
int bs_right(int b) {
  int bas = 1, son = bef_size - sorted[b] + 1;
  while (bas < son) {
    int orta = bas + son >> 1;
    int tt = ctr(b, orta, 1);
    if (!tt)
      bas = orta + 1;
    else
      son = orta;
  }
  return bas;
}
int main() {
  cin >> s;
  scanf("%d", &q);
  s = '0' + s;
  int ll = s.size();
  bef_size = s.size() - 1;
  for (int i = 1; i <= q; i++) {
    cin >> ss;
    scanf("%d %d", &a[i][0], &a[i][1]);
    int temp = ss.size();
    w[ll] = -1;
    for (int j = ll + 1; j <= ll + temp; j++) w[j] = i;
    s = s + '$' + ss;
    ll = s.size();
  }
  n = ll - 1;
  for (int i = 1; i <= n; i++) suff[i] = s[i];
  for (int i = 1; i <= n; i++) LOG[i] = log2(i);
  for (int j = 1; j <= logN; j++) {
    for (int i = 1; i <= n; i++)
      C[i] =
          make_pair(make_pair(suff[i], suff[min(i + (1 << j - 1), n + 1)]), i);
    sort(C + 1, C + n + 1);
    for (int i = 1; i <= n; i++)
      suff[C[i].second] =
          suff[C[i - 1].second] + (C[i - 1].first != C[i].first);
  }
  for (int i = 1; i <= n; i++) sorted[suff[i]] = i;
  int j = 0;
  for (int i = 1; i <= n; i++) {
    if (suff[i] == 1) continue;
    while (i + j <= n && sorted[suff[i] - 1] + j <= n &&
           s[sorted[suff[i] - 1] + j] == s[i + j])
      j++;
    rmq[suff[i] - 1][0] = j;
    if (j) j--;
  }
  rmq[n][0] = 1000000000;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= q; j++) {
      dp[i][j] = dp[i - 1][j] + (w[sorted[i]] == j);
      if (w[sorted[i]] == j)
        bef[i][j] = i;
      else
        bef[i][j] = bef[i - 1][j];
    }
  for (int j = 1; j <= logN; j++)
    for (int i = 1; i <= n; i++)
      rmq[i][j] = min(rmq[i][j - 1], rmq[min(n, i + (1 << j - 1))][j - 1]);
  int r = 0, l = 0, bef = 0;
  for (int i = 1; i <= n; i++) {
    if (w[sorted[i]] != 0) continue;
    l = bs_left(i);
    if (bef)
      r = max(bs_right(i), query(bef, i) + 1);
    else
      r = bs_right(i);
    bef = i;
    if (ctr(i, l, 0) && ctr(i, r, 1)) ans += max(l - r + 1, 0);
  }
  cout << ans << '\n';
  return 0;
}

```