#include <bits/stdc++.h>
using namespace std;
int n;
int l[600006], r[600006];
struct Suffix_Auto_Machine {
  struct state {
    int to[27], len, pre, endpos[11];
  } s[600006];
  struct apple {
    int to, next;
  } indexx[600006];
  int head[600006], cnt;
  void init(int x, int y) {
    indexx[++cnt].to = y;
    indexx[cnt].next = head[x];
    head[x] = cnt;
  }
  int fin;
  int siz;
  int ans;
  void res() {
    fin = ++siz;
    return;
  }
  void insert(int cc, int id) {
    int nwp, nwq, lsp, lsq;
    if (s[fin].to[cc]) {
      lsp = fin;
      lsq = s[lsp].to[cc];
      if (s[lsq].len == s[lsp].len + 1) {
        fin = lsq;
      } else {
        nwq = ++siz;
        s[nwq] = s[lsq];
        memset(s[nwq].endpos, 0, sizeof(s[nwq].endpos));
        s[nwq].len = s[lsp].len + 1;
        s[lsq].pre = nwq;
        while (s[lsp].to[cc] == lsq) {
          s[lsp].to[cc] = nwq;
          lsp = s[lsp].pre;
        }
        fin = nwq;
      }
    } else {
      nwp = ++siz;
      s[nwp].len = s[fin].len + 1;
      for (lsp = fin; !s[lsp].to[cc]; lsp = s[lsp].pre) {
        s[lsp].to[cc] = nwp;
      }
      if (!lsp) {
        s[nwp].pre = 1;
      } else {
        lsq = s[lsp].to[cc];
        if (s[lsq].len == s[lsp].len + 1) {
          s[nwp].pre = lsq;
        } else {
          nwq = ++siz;
          s[nwq] = s[lsq];
          memset(s[nwq].endpos, 0, sizeof(s[nwq].endpos));
          s[nwq].len = s[lsp].len + 1;
          s[nwp].pre = s[lsq].pre = nwq;
          while (s[lsp].to[cc] == lsq) {
            s[lsp].to[cc] = nwq;
            lsp = s[lsp].pre;
          }
        }
      }
      fin = nwp;
    }
    s[fin].endpos[id]++;
    return;
  }
  void build() {
    for (int i = 1; i <= siz; i++) {
      init(s[i].pre, i);
    }
  }
  void dfs(int u) {
    for (int i = head[u]; i; i = indexx[i].next) {
      int v = indexx[i].to;
      dfs(v);
      for (int j = 0; j <= n; j++) {
        s[u].endpos[j] += s[v].endpos[j];
      }
    }
    if (u) {
      if (s[u].endpos[0]) {
        int flag = 1;
        for (int j = 1; j <= n; j++) {
          flag &= (s[u].endpos[j] >= l[j] && s[u].endpos[j] <= r[j]);
        }
        ans += flag * (s[u].len - s[s[u].pre].len);
      }
    }
  }
} sam;
char s[600006];
int len;
int main() {
  sam.res();
  scanf("%s", s + 1);
  len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    sam.insert(s[i] - 'a' + 1, 0);
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    sam.fin = 1;
    scanf("%s", s + 1);
    len = strlen(s + 1);
    for (int j = 1; j <= len; j++) {
      sam.insert(s[j] - 'a' + 1, i);
    }
    scanf("%d%d", &l[i], &r[i]);
  }
  sam.build();
  sam.dfs(1);
  printf("%d\n", sam.ans);
}

### 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;
int n;
int l[600006], r[600006];
struct Suffix_Auto_Machine {
  struct state {
    int to[27], len, pre, endpos[11];
  } s[600006];
  struct apple {
    int to, next;
  } indexx[600006];
  int head[600006], cnt;
  void init(int x, int y) {
    indexx[++cnt].to = y;
    indexx[cnt].next = head[x];
    head[x] = cnt;
  }
  int fin;
  int siz;
  int ans;
  void res() {
    fin = ++siz;
    return;
  }
  void insert(int cc, int id) {
    int nwp, nwq, lsp, lsq;
    if (s[fin].to[cc]) {
      lsp = fin;
      lsq = s[lsp].to[cc];
      if (s[lsq].len == s[lsp].len + 1) {
        fin = lsq;
      } else {
        nwq = ++siz;
        s[nwq] = s[lsq];
        memset(s[nwq].endpos, 0, sizeof(s[nwq].endpos));
        s[nwq].len = s[lsp].len + 1;
        s[lsq].pre = nwq;
        while (s[lsp].to[cc] == lsq) {
          s[lsp].to[cc] = nwq;
          lsp = s[lsp].pre;
        }
        fin = nwq;
      }
    } else {
      nwp = ++siz;
      s[nwp].len = s[fin].len + 1;
      for (lsp = fin; !s[lsp].to[cc]; lsp = s[lsp].pre) {
        s[lsp].to[cc] = nwp;
      }
      if (!lsp) {
        s[nwp].pre = 1;
      } else {
        lsq = s[lsp].to[cc];
        if (s[lsq].len == s[lsp].len + 1) {
          s[nwp].pre = lsq;
        } else {
          nwq = ++siz;
          s[nwq] = s[lsq];
          memset(s[nwq].endpos, 0, sizeof(s[nwq].endpos));
          s[nwq].len = s[lsp].len + 1;
          s[nwp].pre = s[lsq].pre = nwq;
          while (s[lsp].to[cc] == lsq) {
            s[lsp].to[cc] = nwq;
            lsp = s[lsp].pre;
          }
        }
      }
      fin = nwp;
    }
    s[fin].endpos[id]++;
    return;
  }
  void build() {
    for (int i = 1; i <= siz; i++) {
      init(s[i].pre, i);
    }
  }
  void dfs(int u) {
    for (int i = head[u]; i; i = indexx[i].next) {
      int v = indexx[i].to;
      dfs(v);
      for (int j = 0; j <= n; j++) {
        s[u].endpos[j] += s[v].endpos[j];
      }
    }
    if (u) {
      if (s[u].endpos[0]) {
        int flag = 1;
        for (int j = 1; j <= n; j++) {
          flag &= (s[u].endpos[j] >= l[j] && s[u].endpos[j] <= r[j]);
        }
        ans += flag * (s[u].len - s[s[u].pre].len);
      }
    }
  }
} sam;
char s[600006];
int len;
int main() {
  sam.res();
  scanf("%s", s + 1);
  len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    sam.insert(s[i] - 'a' + 1, 0);
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    sam.fin = 1;
    scanf("%s", s + 1);
    len = strlen(s + 1);
    for (int j = 1; j <= len; j++) {
      sam.insert(s[j] - 'a' + 1, i);
    }
    scanf("%d%d", &l[i], &r[i]);
  }
  sam.build();
  sam.dfs(1);
  printf("%d\n", sam.ans);
}

```