#include <bits/stdc++.h>
using namespace std;
const int N = 600005;
const int M = N * 2;
const int S = 50005;
inline int IN() {
  char ch;
  (ch = getchar());
  int f = 0, x = 0;
  for (; ((ch) == '\n' || (ch) == '\r' || (ch) == '\t' || (ch) == ' ');
       (ch = getchar()))
    ;
  if (ch == '-') f = 1, (ch = getchar());
  for (; !((ch) == '\n' || (ch) == '\r' || (ch) == '\t' || (ch) == ' ');
       (ch = getchar()))
    x = x * 10 + ch - '0';
  return (f) ? (-x) : (x);
}
int n, m, l[16], r[16];
char s[S];
long long ans;
int last, dex, root, dp[11][M];
int tree[M][38], step[M];
int pre[M], vis[M];
struct SAM {
  void init() {
    dex = root = last = 1;
    memset(tree, 0, sizeof(tree));
    memset(step, 0, sizeof(step));
    memset(pre, 0, sizeof(pre));
  }
  void ext(int x) {
    int p, q = last, skydec = (++dex);
    step[skydec] = step[q] + 1;
    for (p = q; p && !tree[p][x]; p = pre[p]) tree[p][x] = skydec;
    if (!p)
      pre[skydec] = 1;
    else {
      int gt = tree[p][x];
      if (step[p] + 1 == step[gt])
        pre[skydec] = gt;
      else {
        int nq = (++dex);
        step[nq] = step[p] + 1;
        pre[nq] = pre[gt];
        pre[gt] = pre[skydec] = nq;
        memcpy(tree[nq], tree[gt], sizeof(tree[gt]));
        for (int z = p; z && tree[z][x] == gt; z = pre[z]) tree[z][x] = nq;
      }
    }
    last = skydec;
  }
  void dfs(int u) {
    if (vis[u]) return;
    vis[u] = 1;
    for (int i = (int)0; i <= (int)m; i++)
      if (tree[u][26 + i]) dp[i][u] = 1;
    for (int i = (int)0; i <= (int)25; i++) {
      int v = tree[u][i];
      if (!v) continue;
      dfs(v);
      for (int x = (int)0; x <= (int)m; x++) dp[x][u] += dp[x][v];
    }
    bool flag = 1;
    for (int x = (int)0; x <= (int)m; x++)
      if (!(l[x] <= dp[x][u] && dp[x][u] <= r[x])) flag = 0;
    if (flag) ans += step[u] - step[pre[u]];
  }
} sam;
int main() {
  sam.init();
  l[0] = 1;
  r[0] = 10000005;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = (int)1; i <= (int)n; i++) sam.ext(s[i] - 'a');
  sam.ext(26);
  m = IN();
  for (int v = (int)1; v <= (int)m; v++) {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for (int i = (int)1; i <= (int)n; i++) sam.ext(s[i] - 'a');
    sam.ext(26 + v);
    l[v] = IN(), r[v] = IN();
  }
  sam.dfs(1);
  printf("%I64d\n", ans);
  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 = 600005;
const int M = N * 2;
const int S = 50005;
inline int IN() {
  char ch;
  (ch = getchar());
  int f = 0, x = 0;
  for (; ((ch) == '\n' || (ch) == '\r' || (ch) == '\t' || (ch) == ' ');
       (ch = getchar()))
    ;
  if (ch == '-') f = 1, (ch = getchar());
  for (; !((ch) == '\n' || (ch) == '\r' || (ch) == '\t' || (ch) == ' ');
       (ch = getchar()))
    x = x * 10 + ch - '0';
  return (f) ? (-x) : (x);
}
int n, m, l[16], r[16];
char s[S];
long long ans;
int last, dex, root, dp[11][M];
int tree[M][38], step[M];
int pre[M], vis[M];
struct SAM {
  void init() {
    dex = root = last = 1;
    memset(tree, 0, sizeof(tree));
    memset(step, 0, sizeof(step));
    memset(pre, 0, sizeof(pre));
  }
  void ext(int x) {
    int p, q = last, skydec = (++dex);
    step[skydec] = step[q] + 1;
    for (p = q; p && !tree[p][x]; p = pre[p]) tree[p][x] = skydec;
    if (!p)
      pre[skydec] = 1;
    else {
      int gt = tree[p][x];
      if (step[p] + 1 == step[gt])
        pre[skydec] = gt;
      else {
        int nq = (++dex);
        step[nq] = step[p] + 1;
        pre[nq] = pre[gt];
        pre[gt] = pre[skydec] = nq;
        memcpy(tree[nq], tree[gt], sizeof(tree[gt]));
        for (int z = p; z && tree[z][x] == gt; z = pre[z]) tree[z][x] = nq;
      }
    }
    last = skydec;
  }
  void dfs(int u) {
    if (vis[u]) return;
    vis[u] = 1;
    for (int i = (int)0; i <= (int)m; i++)
      if (tree[u][26 + i]) dp[i][u] = 1;
    for (int i = (int)0; i <= (int)25; i++) {
      int v = tree[u][i];
      if (!v) continue;
      dfs(v);
      for (int x = (int)0; x <= (int)m; x++) dp[x][u] += dp[x][v];
    }
    bool flag = 1;
    for (int x = (int)0; x <= (int)m; x++)
      if (!(l[x] <= dp[x][u] && dp[x][u] <= r[x])) flag = 0;
    if (flag) ans += step[u] - step[pre[u]];
  }
} sam;
int main() {
  sam.init();
  l[0] = 1;
  r[0] = 10000005;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = (int)1; i <= (int)n; i++) sam.ext(s[i] - 'a');
  sam.ext(26);
  m = IN();
  for (int v = (int)1; v <= (int)m; v++) {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for (int i = (int)1; i <= (int)n; i++) sam.ext(s[i] - 'a');
    sam.ext(26 + v);
    l[v] = IN(), r[v] = IN();
  }
  sam.dfs(1);
  printf("%I64d\n", ans);
  return 0;
}

```