#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const int N = 500100;
char s[N], s1[50100];
int wa[N], wb[N], wc[N], sa[N], ht[N], rankk[N], rmq[N][21];
int tag[N], sum[N][12], ll[N], rr[N], up[N], down[N], LOG[N];
int n, q, n1;
long long out = 0;
void getsa() {
  int i, j, *x = wa, *y = wc, *t, m = 200, p = 0;
  for (i = 0; i <= m; i++) wb[i] = 0;
  for (i = 1; i <= n; i++) wb[x[i] = s[i]]++;
  for (i = 1; i <= m; i++) wb[i] += wb[i - 1];
  for (i = n; i; --i) sa[wb[x[i]]--] = i;
  for (j = 1; p < n; j *= 2, m = p) {
    p = 0;
    for (i = n - j + 1; i <= n; i++) y[++p] = i;
    for (i = 1; i <= n; i++)
      if (sa[i] > j) y[++p] = sa[i] - j;
    for (i = 0; i <= m; i++) wb[i] = 0;
    for (i = 1; i <= n; i++) wb[x[y[i]]]++;
    for (i = 1; i <= m; i++) wb[i] += wb[i - 1];
    for (i = n; i; --i) sa[wb[x[y[i]]]--] = y[i];
    t = x;
    x = y;
    y = t;
    p = 1;
    x[sa[1]] = 1;
    for (i = 2; i <= n; i++)
      if ((y[sa[i]] == y[sa[i - 1]]) && (y[sa[i] + j] == y[sa[i - 1] + j]))
        x[sa[i]] = p;
      else
        x[sa[i]] = ++p;
  }
}
void geth() {
  int i, j, k = 0;
  for (i = 1; i <= n; i++) rankk[sa[i]] = i;
  for (i = 1; i <= n; i++) {
    if (k) k--;
    if (rankk[i] == 1) continue;
    j = sa[rankk[i] - 1];
    while (s[j + k] == s[i + k]) k++;
    ht[rankk[i]] = k;
  }
}
void ycl() {
  int i, j;
  for (i = 1; i <= n; i++)
    for (j = 2; j <= q + 1; j++) sum[i][j] = sum[i - 1][j] + (tag[sa[i]] == j);
  for (j = 0; j <= 19; j++)
    for (i = 1; i + (1 << j) - 1 <= n; i++) rmq[i][j] = inf;
  for (i = 1; i <= n; i++) rmq[i][0] = ht[i];
  for (j = 1; j <= 19; j++)
    for (i = 1; i + (1 << j) - 1 <= n; i++)
      rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << (j - 1))][j - 1]);
  LOG[1] = 0;
  for (i = 2; i <= n; i++) LOG[i] = LOG[i >> 1] + 1;
}
int ask(int l, int r) {
  l = rankk[l];
  r = rankk[r];
  if (l > r) swap(l, r);
  l++;
  int t = LOG[r - l + 1];
  return min(rmq[l][t], rmq[r - (1 << t) + 1][t]);
}
int bin1(int len, int pos) {
  int l, r, mid, ans;
  pos = rankk[pos];
  ans = pos;
  r = pos - 1;
  l = 1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (ask(sa[mid], sa[pos]) >= len)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  return ans;
}
int bin2(int len, int pos) {
  int l, r, mid, ans = pos;
  pos = rankk[pos];
  ans = pos;
  l = pos + 1;
  r = n;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (ask(sa[mid], sa[pos]) >= len)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return ans;
}
bool jud1(int l, int r) {
  for (int i = 2; i <= q + 1; i++)
    if (sum[r][i] - sum[l - 1][i] > up[i]) return 0;
  return 1;
}
bool jud2(int l, int r) {
  for (int i = 2; i <= q + 1; i++)
    if (sum[r][i] - sum[l - 1][i] < down[i]) return 0;
  return 1;
}
int main() {
  int i, j, k, len, l, r, L, R, mid, pre, ans1, ans2;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  n1 = n;
  for (i = 1; i <= n; i++) tag[i] = 1;
  s[++n] = 'A';
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%s", s1 + 1);
    len = strlen(s1 + 1);
    scanf("%d%d", &down[i + 1], &up[i + 1]);
    for (j = 1; j <= len; j++) {
      s[n + j] = s1[j];
      tag[n + j] = i + 1;
    }
    n += len + 1;
    s[n] = 'A' + i;
  }
  getsa();
  geth();
  ycl();
  for (i = 1; i <= n1; i++) {
    l = 1;
    r = n1 - i + 1;
    ans1 = n1;
    while (l <= r) {
      mid = (l + r) >> 1;
      L = bin1(mid, i);
      R = bin2(mid, i);
      if (jud1(L, R))
        ans1 = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    ll[i] = ans1;
    l = 1;
    r = n1 - i + 1;
    ans2 = 0;
    while (l <= r) {
      mid = (l + r) >> 1;
      L = bin1(mid, i);
      R = bin2(mid, i);
      if (jud2(L, R))
        ans2 = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    rr[i] = ans2;
  }
  pre = 0;
  out = 0;
  for (i = 1; i <= n; i++) {
    if (tag[sa[i]] != 1) continue;
    out += max(0, rr[sa[i]] - max(ll[sa[i]] - 1, ask(sa[i], pre)));
    pre = sa[i];
  }
  cout << out << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 inf = 1e9 + 7;
const int N = 500100;
char s[N], s1[50100];
int wa[N], wb[N], wc[N], sa[N], ht[N], rankk[N], rmq[N][21];
int tag[N], sum[N][12], ll[N], rr[N], up[N], down[N], LOG[N];
int n, q, n1;
long long out = 0;
void getsa() {
  int i, j, *x = wa, *y = wc, *t, m = 200, p = 0;
  for (i = 0; i <= m; i++) wb[i] = 0;
  for (i = 1; i <= n; i++) wb[x[i] = s[i]]++;
  for (i = 1; i <= m; i++) wb[i] += wb[i - 1];
  for (i = n; i; --i) sa[wb[x[i]]--] = i;
  for (j = 1; p < n; j *= 2, m = p) {
    p = 0;
    for (i = n - j + 1; i <= n; i++) y[++p] = i;
    for (i = 1; i <= n; i++)
      if (sa[i] > j) y[++p] = sa[i] - j;
    for (i = 0; i <= m; i++) wb[i] = 0;
    for (i = 1; i <= n; i++) wb[x[y[i]]]++;
    for (i = 1; i <= m; i++) wb[i] += wb[i - 1];
    for (i = n; i; --i) sa[wb[x[y[i]]]--] = y[i];
    t = x;
    x = y;
    y = t;
    p = 1;
    x[sa[1]] = 1;
    for (i = 2; i <= n; i++)
      if ((y[sa[i]] == y[sa[i - 1]]) && (y[sa[i] + j] == y[sa[i - 1] + j]))
        x[sa[i]] = p;
      else
        x[sa[i]] = ++p;
  }
}
void geth() {
  int i, j, k = 0;
  for (i = 1; i <= n; i++) rankk[sa[i]] = i;
  for (i = 1; i <= n; i++) {
    if (k) k--;
    if (rankk[i] == 1) continue;
    j = sa[rankk[i] - 1];
    while (s[j + k] == s[i + k]) k++;
    ht[rankk[i]] = k;
  }
}
void ycl() {
  int i, j;
  for (i = 1; i <= n; i++)
    for (j = 2; j <= q + 1; j++) sum[i][j] = sum[i - 1][j] + (tag[sa[i]] == j);
  for (j = 0; j <= 19; j++)
    for (i = 1; i + (1 << j) - 1 <= n; i++) rmq[i][j] = inf;
  for (i = 1; i <= n; i++) rmq[i][0] = ht[i];
  for (j = 1; j <= 19; j++)
    for (i = 1; i + (1 << j) - 1 <= n; i++)
      rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << (j - 1))][j - 1]);
  LOG[1] = 0;
  for (i = 2; i <= n; i++) LOG[i] = LOG[i >> 1] + 1;
}
int ask(int l, int r) {
  l = rankk[l];
  r = rankk[r];
  if (l > r) swap(l, r);
  l++;
  int t = LOG[r - l + 1];
  return min(rmq[l][t], rmq[r - (1 << t) + 1][t]);
}
int bin1(int len, int pos) {
  int l, r, mid, ans;
  pos = rankk[pos];
  ans = pos;
  r = pos - 1;
  l = 1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (ask(sa[mid], sa[pos]) >= len)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  return ans;
}
int bin2(int len, int pos) {
  int l, r, mid, ans = pos;
  pos = rankk[pos];
  ans = pos;
  l = pos + 1;
  r = n;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (ask(sa[mid], sa[pos]) >= len)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return ans;
}
bool jud1(int l, int r) {
  for (int i = 2; i <= q + 1; i++)
    if (sum[r][i] - sum[l - 1][i] > up[i]) return 0;
  return 1;
}
bool jud2(int l, int r) {
  for (int i = 2; i <= q + 1; i++)
    if (sum[r][i] - sum[l - 1][i] < down[i]) return 0;
  return 1;
}
int main() {
  int i, j, k, len, l, r, L, R, mid, pre, ans1, ans2;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  n1 = n;
  for (i = 1; i <= n; i++) tag[i] = 1;
  s[++n] = 'A';
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%s", s1 + 1);
    len = strlen(s1 + 1);
    scanf("%d%d", &down[i + 1], &up[i + 1]);
    for (j = 1; j <= len; j++) {
      s[n + j] = s1[j];
      tag[n + j] = i + 1;
    }
    n += len + 1;
    s[n] = 'A' + i;
  }
  getsa();
  geth();
  ycl();
  for (i = 1; i <= n1; i++) {
    l = 1;
    r = n1 - i + 1;
    ans1 = n1;
    while (l <= r) {
      mid = (l + r) >> 1;
      L = bin1(mid, i);
      R = bin2(mid, i);
      if (jud1(L, R))
        ans1 = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    ll[i] = ans1;
    l = 1;
    r = n1 - i + 1;
    ans2 = 0;
    while (l <= r) {
      mid = (l + r) >> 1;
      L = bin1(mid, i);
      R = bin2(mid, i);
      if (jud2(L, R))
        ans2 = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    rr[i] = ans2;
  }
  pre = 0;
  out = 0;
  for (i = 1; i <= n; i++) {
    if (tag[sa[i]] != 1) continue;
    out += max(0, rr[sa[i]] - max(ll[sa[i]] - 1, ask(sa[i], pre)));
    pre = sa[i];
  }
  cout << out << endl;
  return 0;
}

```