#include <bits/stdc++.h>
using namespace std;
const int N = 2010;
const int M = 2000100;
int m, l[N], r[N], pos[N], len[N], w[M], mask[2][M];
char p[11][N];
bool vis[M], ok[N][N];
map<int, int> trie[2][M];
void f(int at) {
  int n = strlen(p[at]);
  int r = min(1, at);
  len[r] = 1;
  mask[r][0] = 0;
  trie[r][0].clear();
  for (int i = 0; i < n; i++) {
    int now = 0;
    for (int j = i; j < n; j++) {
      int ch = p[at][j] - 'a';
      if ((mask[r][now] & (1 << ch)) == 0) {
        mask[r][len[r]] = 0;
        trie[r][len[r]].clear();
        w[len[r]] = 0;
        mask[r][now] |= (1 << ch);
        trie[r][now][ch] = len[r]++;
      }
      now = trie[r][now][ch];
      w[now]++;
    }
  }
}
int main() {
  scanf("%s %d", p[0], &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s %d %d", p[i], &l[i], &r[i]);
  }
  int n = strlen(p[0]);
  memset(ok, true, sizeof ok);
  for (int k = 1; k <= m; k++) {
    f(k);
    for (int i = 0; i < n; i++) {
      pos[k] = 0;
      for (int j = i; j < n; j++) {
        int ch = p[0][j] - 'a';
        int where = pos[k];
        if (where == -1 || trie[1][where].count(ch) == 0) {
          pos[k] = -1;
        } else {
          pos[k] = trie[1][where][ch];
        }
        int many = 0;
        if (pos[k] != -1) {
          many += w[pos[k]];
        }
        if (many < l[k] || many > r[k]) {
          ok[i][j] = false;
        }
      }
    }
    for (int i = 0; i < l[1]; i++) {
      trie[1][i].clear();
    }
  }
  int res = 0;
  f(0);
  memset(vis, false, sizeof vis);
  for (int i = 0; i < n; i++) {
    int now = 0;
    for (int j = i; j < n; j++) {
      int ch = p[0][j] - 'a';
      now = trie[0][now][ch];
      res += ((!vis[now] && ok[i][j]) ? 1 : 0);
      vis[now] = true;
    }
  }
  printf("%d\n", res);
  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>
using namespace std;
const int N = 2010;
const int M = 2000100;
int m, l[N], r[N], pos[N], len[N], w[M], mask[2][M];
char p[11][N];
bool vis[M], ok[N][N];
map<int, int> trie[2][M];
void f(int at) {
  int n = strlen(p[at]);
  int r = min(1, at);
  len[r] = 1;
  mask[r][0] = 0;
  trie[r][0].clear();
  for (int i = 0; i < n; i++) {
    int now = 0;
    for (int j = i; j < n; j++) {
      int ch = p[at][j] - 'a';
      if ((mask[r][now] & (1 << ch)) == 0) {
        mask[r][len[r]] = 0;
        trie[r][len[r]].clear();
        w[len[r]] = 0;
        mask[r][now] |= (1 << ch);
        trie[r][now][ch] = len[r]++;
      }
      now = trie[r][now][ch];
      w[now]++;
    }
  }
}
int main() {
  scanf("%s %d", p[0], &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s %d %d", p[i], &l[i], &r[i]);
  }
  int n = strlen(p[0]);
  memset(ok, true, sizeof ok);
  for (int k = 1; k <= m; k++) {
    f(k);
    for (int i = 0; i < n; i++) {
      pos[k] = 0;
      for (int j = i; j < n; j++) {
        int ch = p[0][j] - 'a';
        int where = pos[k];
        if (where == -1 || trie[1][where].count(ch) == 0) {
          pos[k] = -1;
        } else {
          pos[k] = trie[1][where][ch];
        }
        int many = 0;
        if (pos[k] != -1) {
          many += w[pos[k]];
        }
        if (many < l[k] || many > r[k]) {
          ok[i][j] = false;
        }
      }
    }
    for (int i = 0; i < l[1]; i++) {
      trie[1][i].clear();
    }
  }
  int res = 0;
  f(0);
  memset(vis, false, sizeof vis);
  for (int i = 0; i < n; i++) {
    int now = 0;
    for (int j = i; j < n; j++) {
      int ch = p[0][j] - 'a';
      now = trie[0][now][ch];
      res += ((!vis[now] && ok[i][j]) ? 1 : 0);
      vis[now] = true;
    }
  }
  printf("%d\n", res);
  return 0;
}

```