#include <bits/stdc++.h>
using namespace std;
const int maxn = 800005, maxk = 27, maxm = 15, maxe = maxn;
int i, j, k, m, n, tot, last, e;
int link[maxn], nxt[maxn][maxk], len[maxn], size[maxn][maxm], l[maxn], r[maxn],
    start[maxn], to[maxe], then[maxe];
string s, T;
long long ans;
inline void add(int x, int y) { then[++e] = start[x], start[x] = e, to[e] = y; }
int extend(int last, int x, int id) {
  int now = ++tot, pos = last, tmp, cln;
  size[now][id]++;
  len[now] = len[pos] + 1;
  while (pos != 0 && nxt[pos][x] == 0) nxt[pos][x] = now, pos = link[pos];
  if (pos == 0) {
    link[now] = 1;
    return now;
  }
  tmp = nxt[pos][x];
  if (len[tmp] == len[pos] + 1) {
    link[now] = tmp;
    return now;
  }
  cln = ++tot;
  len[cln] = len[pos] + 1;
  for (int i = 1; i <= 26; i++) nxt[cln][i] = nxt[tmp][i];
  link[cln] = link[tmp], link[tmp] = link[now] = cln;
  while (pos != 0 && nxt[pos][x] == tmp) nxt[pos][x] = cln, pos = link[pos];
  return now;
}
void dfs(int x) {
  for (int i = start[x]; i; i = then[i]) {
    int y = to[i];
    dfs(y);
    for (int j = 0; j <= n; j++) size[x][j] += size[y][j];
  }
  for (int i = 0; i <= n; i++)
    if (size[x][i] < l[i] || size[x][i] > r[i]) return;
  ans += len[x] - len[link[x]];
}
int main() {
  last = tot = 1;
  cin >> T;
  for (int i = 0; i < T.size(); i++) last = extend(last, T[i] - 96, 0);
  scanf("%d", &n);
  l[0] = 1, r[0] = 1000000000;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    scanf("%d%d", &l[i], &r[i]);
    last = 1;
    for (int j = 0; j < s.size(); j++) last = extend(last, s[j] - 96, i);
  }
  for (int i = 2; i <= tot; i++) add(link[i], i);
  dfs(1);
  printf("%lld\n", ans);
  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 maxn = 800005, maxk = 27, maxm = 15, maxe = maxn;
int i, j, k, m, n, tot, last, e;
int link[maxn], nxt[maxn][maxk], len[maxn], size[maxn][maxm], l[maxn], r[maxn],
    start[maxn], to[maxe], then[maxe];
string s, T;
long long ans;
inline void add(int x, int y) { then[++e] = start[x], start[x] = e, to[e] = y; }
int extend(int last, int x, int id) {
  int now = ++tot, pos = last, tmp, cln;
  size[now][id]++;
  len[now] = len[pos] + 1;
  while (pos != 0 && nxt[pos][x] == 0) nxt[pos][x] = now, pos = link[pos];
  if (pos == 0) {
    link[now] = 1;
    return now;
  }
  tmp = nxt[pos][x];
  if (len[tmp] == len[pos] + 1) {
    link[now] = tmp;
    return now;
  }
  cln = ++tot;
  len[cln] = len[pos] + 1;
  for (int i = 1; i <= 26; i++) nxt[cln][i] = nxt[tmp][i];
  link[cln] = link[tmp], link[tmp] = link[now] = cln;
  while (pos != 0 && nxt[pos][x] == tmp) nxt[pos][x] = cln, pos = link[pos];
  return now;
}
void dfs(int x) {
  for (int i = start[x]; i; i = then[i]) {
    int y = to[i];
    dfs(y);
    for (int j = 0; j <= n; j++) size[x][j] += size[y][j];
  }
  for (int i = 0; i <= n; i++)
    if (size[x][i] < l[i] || size[x][i] > r[i]) return;
  ans += len[x] - len[link[x]];
}
int main() {
  last = tot = 1;
  cin >> T;
  for (int i = 0; i < T.size(); i++) last = extend(last, T[i] - 96, 0);
  scanf("%d", &n);
  l[0] = 1, r[0] = 1000000000;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    scanf("%d%d", &l[i], &r[i]);
    last = 1;
    for (int j = 0; j < s.size(); j++) last = extend(last, s[j] - 96, i);
  }
  for (int i = 2; i <= tot; i++) add(link[i], i);
  dfs(1);
  printf("%lld\n", ans);
  return 0;
}

```