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 long long inf = (long long)1e18;
const long long mod = (long long)1e9 + 7;
const double eps = (double)1e-9;
const double pi = acos(-1.0);
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};
const int N = 1000500;
long long ans;
char s[N];
vector<pair<int, int> > pr[N];
int n, cnt, lp[N], f[N];
inline void add(int i) {
  for (auto x : pr[i]) {
    int was = f[x.first];
    f[x.first] += x.second;
    int now = f[x.first];
    if (was >= 0 && now < 0) cnt++;
    if (was < 0 && now >= 0) cnt--;
  }
}
inline void del(int i) {
  for (auto x : pr[i]) {
    int was = f[x.first];
    f[x.first] -= x.second;
    int now = f[x.first];
    if (was >= 0 && now < 0) cnt++;
    if (was < 0 && now >= 0) cnt--;
  }
}
int main() {
  for (int i = 2; i * i < N; i++) {
    if (!lp[i]) {
      for (int j = i * i; j < N; j += i)
        if (!lp[j]) lp[j] = i;
    }
  }
  for (int i = 2; i < N; i++)
    if (!lp[i]) lp[i] = i;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    while (x > 1) {
      int p = lp[x], pw = 0;
      while (x % p == 0) {
        pw++;
        x /= p;
      }
      pr[i].push_back({p, pw});
    }
  }
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) {
    if (s[i] != '/') continue;
    for (auto &j : pr[i]) j.second = -j.second;
  }
  for (int i = 1, ptr = 1; i <= n; i++) {
    ptr = max(ptr, i);
    while (ptr <= n) {
      add(ptr);
      if (cnt > 0) {
        del(ptr);
        break;
      }
      ptr++;
    }
    ans += ptr - i;
    if (i < ptr) del(i);
  }
  printf("%lld", ans);
  return 0;
}
