#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
char s[N];
int n, a[N], vis[N], nxt[N];
vector<int> st[N];
void init() {
  vector<int> p;
  for (int i = 2; i < N; i++) {
    if (!vis[i]) vis[i] = i, p.push_back(i);
    for (auto j : p) {
      if (i * j >= N) break;
      vis[i * j] = j;
      if (i % j == 0) break;
    }
  }
}
void add(int x, int p) {
  while (x != 1) {
    int w = vis[x];
    st[w].push_back(p);
    x /= w;
  }
}
void del(int x, int p) {
  while (x != 1) {
    int w = vis[x];
    if (st[w].empty()) return void(nxt[p] = 0);
    nxt[p] = min(nxt[p], st[w].back());
    st[w].pop_back();
    x /= w;
  }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) {
    nxt[i] = i;
    if (s[i] == '*')
      add(a[i], i);
    else
      del(a[i], i);
  }
  vector<pair<int, int>> f;
  long long res = 0;
  for (int i = n; i >= 1; i--) {
    long long cnt = 1;
    while (!f.empty() && f.back().first >= nxt[i]) {
      cnt += f.back().second;
      f.pop_back();
    }
    f.push_back({nxt[i], cnt});
    if (nxt[i] == i) res += cnt;
  }
  printf("%lld\n", res);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1e6 + 5;
char s[N];
int n, a[N], vis[N], nxt[N];
vector<int> st[N];
void init() {
  vector<int> p;
  for (int i = 2; i < N; i++) {
    if (!vis[i]) vis[i] = i, p.push_back(i);
    for (auto j : p) {
      if (i * j >= N) break;
      vis[i * j] = j;
      if (i % j == 0) break;
    }
  }
}
void add(int x, int p) {
  while (x != 1) {
    int w = vis[x];
    st[w].push_back(p);
    x /= w;
  }
}
void del(int x, int p) {
  while (x != 1) {
    int w = vis[x];
    if (st[w].empty()) return void(nxt[p] = 0);
    nxt[p] = min(nxt[p], st[w].back());
    st[w].pop_back();
    x /= w;
  }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) {
    nxt[i] = i;
    if (s[i] == '*')
      add(a[i], i);
    else
      del(a[i], i);
  }
  vector<pair<int, int>> f;
  long long res = 0;
  for (int i = n; i >= 1; i--) {
    long long cnt = 1;
    while (!f.empty() && f.back().first >= nxt[i]) {
      cnt += f.back().second;
      f.pop_back();
    }
    f.push_back({nxt[i], cnt});
    if (nxt[i] == i) res += cnt;
  }
  printf("%lld\n", res);
  return 0;
}

```