#include <bits/stdc++.h>
using namespace std;
const int N = 1000000;
bool primeflag[N + 9];
const int CP = 78498;
int Prime[CP], tot, fac[N + 9];
void prework() {
  for (int i = 2; i <= N; i++) {
    if (!primeflag[i]) Prime[tot] = i, fac[i] = tot, tot++;
    for (int j = 0; j < tot; j++) {
      if (i * Prime[j] > N) break;
      primeflag[i * Prime[j]] = 1, fac[i * Prime[j]] = j;
      if (i % Prime[j] == 0) break;
    }
  }
}
stack<int> stk[CP];
int n, a[N + 9];
char op[N + 9];
int pre[N + 9];
long long Ans;
struct segment {
  int l, r, val;
};
stack<segment> g;
int main() {
  prework();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  scanf("%s", op + 1);
  for (int i = 1; i <= n; i++) {
    pre[i] = i;
    if (op[i] == '*')
      while (a[i] != 1) stk[fac[a[i]]].push(i), a[i] /= Prime[fac[a[i]]];
    else if (op[i] == '/')
      while (a[i] != 1) {
        if (stk[fac[a[i]]].empty())
          pre[i] = 0;
        else
          pre[i] = min(pre[i], stk[fac[a[i]]].top()), stk[fac[a[i]]].pop();
        a[i] /= Prime[fac[a[i]]];
      }
  }
  for (int i = n; i >= 1; i--) {
    segment now = (segment){i, i, pre[i]};
    while (!g.empty()) {
      segment nxt = g.top();
      if (nxt.val >= pre[i])
        now.r = nxt.r, g.pop();
      else
        break;
    }
    g.push(now);
    if (now.val == i) Ans += now.r - now.l + 1;
  }
  printf("%lld\n", Ans);
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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;
const int N = 1000000;
bool primeflag[N + 9];
const int CP = 78498;
int Prime[CP], tot, fac[N + 9];
void prework() {
  for (int i = 2; i <= N; i++) {
    if (!primeflag[i]) Prime[tot] = i, fac[i] = tot, tot++;
    for (int j = 0; j < tot; j++) {
      if (i * Prime[j] > N) break;
      primeflag[i * Prime[j]] = 1, fac[i * Prime[j]] = j;
      if (i % Prime[j] == 0) break;
    }
  }
}
stack<int> stk[CP];
int n, a[N + 9];
char op[N + 9];
int pre[N + 9];
long long Ans;
struct segment {
  int l, r, val;
};
stack<segment> g;
int main() {
  prework();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  scanf("%s", op + 1);
  for (int i = 1; i <= n; i++) {
    pre[i] = i;
    if (op[i] == '*')
      while (a[i] != 1) stk[fac[a[i]]].push(i), a[i] /= Prime[fac[a[i]]];
    else if (op[i] == '/')
      while (a[i] != 1) {
        if (stk[fac[a[i]]].empty())
          pre[i] = 0;
        else
          pre[i] = min(pre[i], stk[fac[a[i]]].top()), stk[fac[a[i]]].pop();
        a[i] /= Prime[fac[a[i]]];
      }
  }
  for (int i = n; i >= 1; i--) {
    segment now = (segment){i, i, pre[i]};
    while (!g.empty()) {
      segment nxt = g.top();
      if (nxt.val >= pre[i])
        now.r = nxt.r, g.pop();
      else
        break;
    }
    g.push(now);
    if (now.val == i) Ans += now.r - now.l + 1;
  }
  printf("%lld\n", Ans);
  return 0;
}

```