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
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 6;
char s[N];
vector<int> st[N];
int n, a[N], b[N], nxt[N], vis[N], sum[N], now;
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) {
    st[vis[x]].push_back(p);
    x /= vis[x];
  }
}
int del(int x, int p) {
  int res = p - 1;
  while (x != 1) {
    int w = vis[x];
    if (st[w].empty() || st[w].back() <= now) {
      st[w].clear();
      return -1;
    }
    res = min(res, st[w].back());
    st[w].pop_back();
    x /= vis[x];
  }
  return res - 1;
}
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++) b[i] = (s[i] == '*' || a[i] == 1);
  long long cnt = 0, res = 0;
  for (int i = 1; i <= n; i++) {
    cnt += b[i];
    if (!b[i] || i == n) {
      res += (cnt + 1) * cnt / 2;
      cnt = 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (b[i])
      add(a[i], i), nxt[i] = i - 1;
    else
      nxt[i] = del(a[i], i);
    if (nxt[i] == -1) {
      now = i;
      sum[i] = 0;
    } else
      sum[i] = sum[nxt[i]] + 1;
    if (b[i]) continue;
    int cnt = 1;
    for (int j = i + 1; b[j] && j <= n; j++) cnt++;
    res += 1ll * sum[i] * cnt;
  }
  printf("%lld\n", res);
  return 0;
}
