#include <bits/stdc++.h>
using namespace std;
const int maxs = 5e4 + 5, maxns = 5e5 + 5e4 + 5;
int n;
char S[maxs], T[maxs];
int l[15], r[15];
int read() {
  int x = 0, y = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') y = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * y;
}
int id = 0;
int tot;
int t[maxns][26], pos[maxns];
vector<int> b[maxns];
void ins(int m) {
  int x = 0;
  for (int i = 1; i <= m; i++) {
    int to = T[i] - 'a';
    if (!t[x][to]) t[x][to] = ++tot;
    if (id) b[x].push_back(id);
    x = t[x][to];
  }
  if (id) b[x].push_back(id);
  return;
}
int nod = 1;
int len[2 * maxns], nxt[2 * maxns];
int to[2 * maxns][26];
int ed[2 * maxns][11];
int extend(int lst, int c, int k) {
  int p = lst, u = ++nod;
  len[u] = len[p] + 1;
  for (int i = 0; i < b[k].size(); i++) ed[u][b[k][i]]++;
  while (p && to[p][c] == 0) {
    to[p][c] = u;
    p = nxt[p];
  }
  if (!p)
    nxt[u] = 1;
  else {
    int d = to[p][c];
    if (len[d] == len[p] + 1)
      nxt[u] = d;
    else {
      int v = ++nod;
      len[v] = len[p] + 1;
      for (int i = 0; i < 26; i++) to[v][i] = to[d][i];
      nxt[v] = nxt[d];
      nxt[d] = v;
      nxt[u] = v;
      while (p && to[p][c] == d) {
        to[p][c] = v;
        p = nxt[p];
      }
    }
  }
  return u;
}
queue<int> q;
void bfs() {
  for (int i = 0; i < 26; i++)
    if (t[0][i]) {
      pos[t[0][i]] = extend(1, i, t[0][i]);
      q.push(t[0][i]);
    }
  while (!q.empty()) {
    int k = q.front();
    q.pop();
    for (int i = 0; i < 26; i++) {
      int tmp = t[k][i];
      if (!tmp) continue;
      pos[tmp] = extend(pos[k], i, tmp);
      q.push(tmp);
    }
  }
  return;
}
struct node {
  int to, nxt;
} e[2 * maxns];
int cnt;
int head[2 * maxns];
void add(int u, int v) {
  e[cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt++;
  return;
}
void dfs(int x) {
  for (int i = head[x]; i != -1; i = e[i].nxt) {
    int tmp = e[i].to;
    dfs(tmp);
    for (int j = 1; j <= n; j++) ed[x][j] += ed[tmp][j];
  }
  return;
}
bool tag[2 * maxns];
int main() {
  scanf("%s", S + 1);
  for (int i = 1; i <= strlen(S + 1); i++) T[i] = S[i];
  ins(strlen(S + 1));
  n = read();
  for (int i = 1; i <= n; i++) {
    scanf("%s", T + 1);
    id = i;
    ins(strlen(T + 1));
    l[i] = read();
    r[i] = read();
  }
  bfs();
  memset(head, -1, sizeof(head));
  for (int i = 2; i <= nod; i++) add(nxt[i], i);
  dfs(1);
  int p = 1, ans = 0;
  for (int i = 1; i <= strlen(S + 1); i++) {
    int c = S[i] - 'a';
    p = to[p][c];
    int q = p;
    while (q && !tag[q]) {
      tag[q] = 1;
      bool flag = 0;
      for (int j = 1; j <= n; j++)
        if (ed[q][j] < l[j] || ed[q][j] > r[j]) {
          flag = 1;
          break;
        }
      if (!flag) ans += len[q] - len[nxt[q]];
      q = nxt[q];
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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 maxs = 5e4 + 5, maxns = 5e5 + 5e4 + 5;
int n;
char S[maxs], T[maxs];
int l[15], r[15];
int read() {
  int x = 0, y = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') y = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * y;
}
int id = 0;
int tot;
int t[maxns][26], pos[maxns];
vector<int> b[maxns];
void ins(int m) {
  int x = 0;
  for (int i = 1; i <= m; i++) {
    int to = T[i] - 'a';
    if (!t[x][to]) t[x][to] = ++tot;
    if (id) b[x].push_back(id);
    x = t[x][to];
  }
  if (id) b[x].push_back(id);
  return;
}
int nod = 1;
int len[2 * maxns], nxt[2 * maxns];
int to[2 * maxns][26];
int ed[2 * maxns][11];
int extend(int lst, int c, int k) {
  int p = lst, u = ++nod;
  len[u] = len[p] + 1;
  for (int i = 0; i < b[k].size(); i++) ed[u][b[k][i]]++;
  while (p && to[p][c] == 0) {
    to[p][c] = u;
    p = nxt[p];
  }
  if (!p)
    nxt[u] = 1;
  else {
    int d = to[p][c];
    if (len[d] == len[p] + 1)
      nxt[u] = d;
    else {
      int v = ++nod;
      len[v] = len[p] + 1;
      for (int i = 0; i < 26; i++) to[v][i] = to[d][i];
      nxt[v] = nxt[d];
      nxt[d] = v;
      nxt[u] = v;
      while (p && to[p][c] == d) {
        to[p][c] = v;
        p = nxt[p];
      }
    }
  }
  return u;
}
queue<int> q;
void bfs() {
  for (int i = 0; i < 26; i++)
    if (t[0][i]) {
      pos[t[0][i]] = extend(1, i, t[0][i]);
      q.push(t[0][i]);
    }
  while (!q.empty()) {
    int k = q.front();
    q.pop();
    for (int i = 0; i < 26; i++) {
      int tmp = t[k][i];
      if (!tmp) continue;
      pos[tmp] = extend(pos[k], i, tmp);
      q.push(tmp);
    }
  }
  return;
}
struct node {
  int to, nxt;
} e[2 * maxns];
int cnt;
int head[2 * maxns];
void add(int u, int v) {
  e[cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt++;
  return;
}
void dfs(int x) {
  for (int i = head[x]; i != -1; i = e[i].nxt) {
    int tmp = e[i].to;
    dfs(tmp);
    for (int j = 1; j <= n; j++) ed[x][j] += ed[tmp][j];
  }
  return;
}
bool tag[2 * maxns];
int main() {
  scanf("%s", S + 1);
  for (int i = 1; i <= strlen(S + 1); i++) T[i] = S[i];
  ins(strlen(S + 1));
  n = read();
  for (int i = 1; i <= n; i++) {
    scanf("%s", T + 1);
    id = i;
    ins(strlen(T + 1));
    l[i] = read();
    r[i] = read();
  }
  bfs();
  memset(head, -1, sizeof(head));
  for (int i = 2; i <= nod; i++) add(nxt[i], i);
  dfs(1);
  int p = 1, ans = 0;
  for (int i = 1; i <= strlen(S + 1); i++) {
    int c = S[i] - 'a';
    p = to[p][c];
    int q = p;
    while (q && !tag[q]) {
      tag[q] = 1;
      bool flag = 0;
      for (int j = 1; j <= n; j++)
        if (ed[q][j] < l[j] || ed[q][j] > r[j]) {
          flag = 1;
          break;
        }
      if (!flag) ans += len[q] - len[nxt[q]];
      q = nxt[q];
    }
  }
  printf("%d\n", ans);
  return 0;
}

```