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;
  char buf[1 << 21], *p1 = buf, *p2 = buf;
  inline int qread() {
    register char c =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++);
    register int x = 0, f = 1;
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
               ? EOF
               : *p1++);
    }
    while (c >= '0' && c <= '9') {
      x = (x << 3) + (x << 1) + c - 48;
      c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
               ? EOF
               : *p1++);
    }
    return x * f;
  }
  int n, a[1000005], b[1000005], pri[1005], pcnt, pre[1000005], stktop,
      f[1000005], res[1000005];
  bool vis[1005];
  vector<pair<int, int> > v[1000005], opt[1000005];
  pair<int, int> stk[1000005];
  inline int GetRoot(int v) {
    if (f[v] == v) return v;
    return f[v] = GetRoot(f[v]);
  }
  inline void Merge(int x, int y) {
    int u = GetRoot(x), v = GetRoot(y);
    if (u != v) {
      f[u] = v;
      return;
    }
  }
  inline char readchar() {
    char c =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++);
    while (c != '/' && c != '*')
      c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
               ? EOF
               : *p1++);
    return c;
  }
  inline void Read() {
    n = qread();
    for (int i = 1; i <= n; i++) a[i] = qread();
    for (int i = 1; i <= n; i++) {
      char c = readchar();
      if (c == '*')
        b[i] = 1;
      else
        b[i] = -1;
    }
  }
  inline void Prefix() {
    for (int i = 2; i <= 1000; i++) {
      if (!vis[i]) {
        pri[++pcnt] = i;
        for (int j = i * i; j <= 1000; j += i) vis[j] = 1;
      }
    }
    for (int i = 1; i <= n; i++) {
      int tmp = a[i];
      for (int j = 1; j <= pcnt && pri[j] * pri[j] <= tmp; j++) {
        if (tmp % pri[j] == 0) {
          int cnt = 0;
          while (tmp % pri[j] == 0) {
            tmp /= pri[j];
            cnt++;
          }
          v[pri[j]].push_back(make_pair(i, cnt * b[i]));
        }
      }
      if (tmp > 1) v[tmp].push_back(make_pair(i, b[i]));
    }
  }
  inline void Solve() {
    for (int i = 1; i <= 1000000; i++) {
      if (v[i].empty()) continue;
      int siz = v[i].size();
      pre[0] = v[i][0].second;
      for (int j = 1; j < siz; j++) pre[j] = pre[j - 1] + v[i][j].second;
      stktop = 1;
      stk[stktop] = make_pair(0xf3f3f3f3, n + 1);
      for (int j = siz - 1; j >= 0; j--) {
        while (stktop && stk[stktop] >= make_pair(pre[j], v[i][j].first))
          stktop--;
        stk[++stktop] = make_pair(pre[j], v[i][j].first);
        int idx = upper_bound(stk + 1, stk + stktop + 1,
                              make_pair(pre[j] - v[i][j].second, 0)) -
                  stk - 1;
        int l;
        if (j == 0)
          l = 1;
        else
          l = v[i][j - 1].first + 1;
        opt[stk[idx].second - 1].push_back(make_pair(l, v[i][j].first));
      }
    }
    for (int i = 1; i <= n + 1; i++) f[i] = i;
    for (int i = 1; i <= n; i++) res[i] = n;
    for (int i = 0; i <= 1000000; i++) {
      int siz = opt[i].size();
      for (int j = 0; j < siz; j++) {
        int l = opt[i][j].first, r = opt[i][j].second;
        for (int k = GetRoot(l); k <= r; k = GetRoot(k + 1)) {
          res[k] = i;
          Merge(k, k + 1);
        }
      }
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) ans += res[i] - i + 1;
    cout << ans << endl;
  }
  int main() {
    Read();
    Prefix();
    Solve();
    return 0;
  }
  