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.
#include <bits/stdc++.h>
using namespace std;
const int N = 11, L = 50010;
int n, tot, l[N], r[N], ll[N];
char s[L], t[N][L];
struct node {
  int len, num;
  node *ch[37], *fa;
  node() {
    len = 0;
    fa = 0;
    for (int i = 0; i < 37; ++i) ch[i] = 0;
  }
} * last, *root, *pool[N * L];
void extend(int c) {
  node *np = pool[++tot] = new node;
  np->num = tot;
  np->len = last->len + 1;
  node *p = last;
  while (p && !p->ch[c]) {
    p->ch[c] = np;
    p = p->fa;
  }
  if (!p) {
    np->fa = root;
  } else {
    node *q = p->ch[c];
    if (q->len != p->len + 1) {
      node *nq = pool[++tot] = new node;
      nq->num = tot;
      nq->len = p->len + 1;
      for (int i = 0; i < 37; ++i) {
        nq->ch[i] = q->ch[i];
      }
      nq->fa = q->fa;
      q->fa = np->fa = nq;
      while (p && p->ch[c] == q) {
        p->ch[c] = nq;
        p = p->fa;
      }
    } else {
      np->fa = q;
    }
  }
  last = np;
}
long long ans;
int dp[N * L][N];
bool vis[N * L];
void dfs(int u) {
  if (vis[u]) return;
  vis[u] = 1;
  for (int i = 0; i <= n; ++i) {
    if (pool[u]->ch[26 + i]) {
      dp[u][i] = 1;
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (pool[u]->ch[i]) {
      int v = pool[u]->ch[i]->num;
      dfs(v);
      for (int j = 0; j <= n; ++j) {
        dp[u][j] += dp[v][j];
      }
    }
  }
  if (u && dp[u][0]) {
    bool ff = 1;
    for (int i = 1; i <= n && ff; ++i) {
      ff = l[i] <= dp[u][i] && dp[u][i] <= r[i];
    }
    if (ff) {
      ans += pool[u]->len - pool[u]->fa->len;
    }
  }
}
int main() {
  last = root = pool[tot] = new node;
  scanf("%s", s);
  int m = strlen(s);
  for (int i = 0; i < m; ++i) {
    extend(s[i] - 'a');
  }
  extend(26);
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s%d%d", t[i], l + i, r + i);
    ll[i] = strlen(t[i]);
    for (int j = 0; j < ll[i]; ++j) {
      extend(t[i][j] - 'a');
    }
    extend(26 + i);
  }
  ans = 0;
  dfs(0);
  printf("%lld\n", ans);
  return 0;
}
