#include <bits/stdc++.h>
using namespace std;
struct qt {
  int val, id;
  qt(int val = 0, int id = 0) : val(val), id(id) {}
  bool operator<(const qt q) const { return val > q.val; }
};
int n, a[1000005], p[1000005], g[1000005], prime[1000005], pt,
    pn = 1e6, mn[1000005], mx[1000005];
int c[1000005], nans[1000005];
char s[1000005];
vector<int> val[1000005];
priority_queue<qt> pq;
void modify(int x, int v, int i) {
  val[x][c[x]] = i;
  c[x] -= v;
  nans[x] = val[x][c[x] - 1];
  pq.push(qt(nans[x], x));
}
int getans() {
  while (!pq.empty() && nans[pq.top().id] != pq.top().val) pq.pop();
  return pq.top().val;
}
int main() {
  p[1] = 1;
  for (int i = 2; i <= pn; i++) {
    if (!p[i]) prime[++pt] = i, g[i] = i;
    for (int j = 1; j <= pt && i * prime[j] <= pn; j++) {
      p[i * prime[j]] = 1;
      g[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%s", s + 1);
  for (int i = n; i > 0; i--) {
    int u = a[i];
    while (u != 1) {
      int v = g[u];
      if (s[i] == '*')
        c[v]--;
      else
        c[v]++;
      mn[v] = min(mn[v], c[v]);
      mx[v] = max(mx[v], c[v]);
      u /= v;
    }
  }
  for (int i = 1; i <= pn; i++)
    if (!p[i]) {
      val[i].resize(mx[i] - mn[i] + 3);
      for (int j = 0; j <= mx[i] - mn[i] + 2; j++) val[i][j] = n + 1;
      c[i] = -mn[i] + 1;
      nans[i] = n + 1;
      pq.push(qt(nans[i], i));
    }
  long long ans = 0;
  for (int i = n; i > 0; i--) {
    int u = a[i];
    while (u != 1) {
      int v = g[u];
      if (s[i] == '*')
        modify(v, 1, i);
      else
        modify(v, -1, i);
      u /= v;
    }
    ans += getans() - i;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

Kuzya started going to school. He was given math homework in which he was given an array a of length n and an array of symbols b of length n, consisting of symbols '*' and '/'.

Let's denote a path of calculations for a segment [l; r] (1 ≤ l ≤ r ≤ n) in the following way: 

  * Let x=1 initially. For every i from l to r we will consequently do the following: if b_i= '*', x=x*a_i, and if b_i= '/', then x=(x)/(a_i). Let's call a path of calculations for the segment [l; r] a list of all x that we got during the calculations (the number of them is exactly r - l + 1). 



For example, let a=[7, 12, 3, 5, 4, 10, 9], b=[/, *, /, /, /, *, *], l=2, r=6, then the path of calculations for that segment is [12, 4, 0.8, 0.2, 2].

Let's call a segment [l;r] simple if the path of calculations for it contains only integer numbers. 

Kuzya needs to find the number of simple segments [l;r] (1 ≤ l ≤ r ≤ n). Since he obviously has no time and no interest to do the calculations for each option, he asked you to write a program to get to find that number!

Input

The first line contains a single integer n (2 ≤ n ≤ 10^6).

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).

The third line contains n symbols without spaces between them — the array b_1, b_2 … b_n (b_i= '/' or b_i= '*' for every 1 ≤ i ≤ n).

Output

Print a single integer — the number of simple segments [l;r].

Examples

Input


3
1 2 3
*/*


Output


2


Input


7
6 4 10 1 2 15 1
*/*/*//


Output


8

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct qt {
  int val, id;
  qt(int val = 0, int id = 0) : val(val), id(id) {}
  bool operator<(const qt q) const { return val > q.val; }
};
int n, a[1000005], p[1000005], g[1000005], prime[1000005], pt,
    pn = 1e6, mn[1000005], mx[1000005];
int c[1000005], nans[1000005];
char s[1000005];
vector<int> val[1000005];
priority_queue<qt> pq;
void modify(int x, int v, int i) {
  val[x][c[x]] = i;
  c[x] -= v;
  nans[x] = val[x][c[x] - 1];
  pq.push(qt(nans[x], x));
}
int getans() {
  while (!pq.empty() && nans[pq.top().id] != pq.top().val) pq.pop();
  return pq.top().val;
}
int main() {
  p[1] = 1;
  for (int i = 2; i <= pn; i++) {
    if (!p[i]) prime[++pt] = i, g[i] = i;
    for (int j = 1; j <= pt && i * prime[j] <= pn; j++) {
      p[i * prime[j]] = 1;
      g[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%s", s + 1);
  for (int i = n; i > 0; i--) {
    int u = a[i];
    while (u != 1) {
      int v = g[u];
      if (s[i] == '*')
        c[v]--;
      else
        c[v]++;
      mn[v] = min(mn[v], c[v]);
      mx[v] = max(mx[v], c[v]);
      u /= v;
    }
  }
  for (int i = 1; i <= pn; i++)
    if (!p[i]) {
      val[i].resize(mx[i] - mn[i] + 3);
      for (int j = 0; j <= mx[i] - mn[i] + 2; j++) val[i][j] = n + 1;
      c[i] = -mn[i] + 1;
      nans[i] = n + 1;
      pq.push(qt(nans[i], i));
    }
  long long ans = 0;
  for (int i = n; i > 0; i--) {
    int u = a[i];
    while (u != 1) {
      int v = g[u];
      if (s[i] == '*')
        modify(v, 1, i);
      else
        modify(v, -1, i);
      u /= v;
    }
    ans += getans() - i;
  }
  printf("%lld\n", ans);
  return 0;
}

```