description: 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
solution: #include <bits/stdc++.h>
  using namespace std;
  using namespace std;
  inline int read() {
    char ch;
    bool sign = 0;
    int res = 0;
    while (!isdigit(ch = getchar()))
      if (ch == '-') sign = 1;
    while (isdigit(ch)) res = res * 10 + (ch ^ 48), ch = getchar();
    if (sign)
      return -res;
    else
      return res;
  }
  const int maxn = 1000000 + 10;
  const int mod = 1e9 + 7;
  const int inf = 1e9;
  const int maxm = 1000 + 10;
  vector<int> prime;
  int cnt, vis[maxn];
  inline void init() {
    for (int i = 2; i < maxn; ++i) {
      if (!vis[i]) {
        cnt++;
        prime.push_back(i);
      }
      for (int j = 0; j < cnt && prime[j] * i < maxn; ++j) {
        vis[prime[j] * i] = 1;
        if (i % prime[j] == 0) break;
      }
    }
  }
  int a[maxn], n, sum[maxn];
  char b[maxn];
  int dp[maxn];
  inline void brute(int p) {
    for (int i = (1); i <= (n); ++i) {
      int now = 0;
      for (; a[i] % p == 0;) now++, a[i] /= p;
      if (b[i] == '/') now *= -1;
      sum[i] = sum[i - 1] + now;
    }
    sum[n + 1] = -inf;
    vector<int> sk;
    sk.push_back(n + 1);
    for (int i = (n); i >= (0); --i) {
      for (; sum[sk.back()] >= sum[i];) sk.pop_back();
      dp[i + 1] = min(dp[i + 1], sk.back() - 1);
      sk.push_back(i);
    }
  }
  vector<int> pos[maxn];
  int has[maxn];
  int f[maxn];
  int pre[maxn];
  int illpos;
  set<int> fval;
  inline long long solve() {
    long long ans = 0;
    for (int i = (1); i <= (n); ++i) dp[i] = n;
    for (int i = (0); i < (cnt); ++i) {
      if (prime[i] * prime[i] <= 1000000) {
        brute(prime[i]);
      } else
        break;
    }
    f[n + 1] = n;
    fval.insert(n + 1);
    illpos = n + 1;
    for (int i = (n); i >= (1); --i) {
      assert(vis[a[i]] == 0);
      if (a[i] == 1) {
        f[i] = f[i + 1];
      } else {
        if (b[i] == '/') {
          has[a[i]]--;
          pos[a[i]].push_back(-i);
          fval.erase(pre[a[i]]);
          pre[a[i]] = i;
          fval.insert(i);
          f[i] = i - 1;
        } else {
          if (!pos[a[i]].empty() && pos[a[i]].back() < 0) {
            pos[a[i]].pop_back();
            fval.erase(pre[a[i]]);
          }
          if (!pos[a[i]].empty() && pos[a[i]].back() < 0) {
            pre[a[i]] = -pos[a[i]].back();
            fval.insert(-pos[a[i]].back());
          }
          f[i] = (*fval.begin()) - 1;
        }
      }
    }
    cerr << f[1] << endl;
    for (int i = (1); i <= (n); ++i) {
      assert(min(f[i], dp[i]) + 1 >= i);
      ans += min(f[i], dp[i]) - i + 1;
    }
    return ans;
  }
  int main(void) {
    init();
    scanf("%d", &(n));
    for (int i = (1); i <= (n); ++i) scanf("%d", &(a[i]));
    scanf("%s", b + 1);
    printf("%lld\n", solve());
    return 0;
  }
  