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>
  long long read() {
    long long x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + (c - '0'), c = getchar();
    return f * x;
  }
  int min(int a, int b) { return a < b ? a : b; }
  int max(int a, int b) { return a > b ? a : b; }
  template <typename T>
  bool umin(T& a, T t) {
    if (a > t) return a = t, 1;
    return 0;
  }
  template <typename T>
  bool umax(T& a, T t) {
    if (a < t) return a = t, 1;
    return 0;
  }
  const int MAXN = 1000011, INF = 1.233e9;
  int pri[MAXN], cnt = 0;
  int minp[MAXN];
  bool vis[MAXN];
  void sieve(int n) {
    vis[1] = 1;
    for (int i = 2; i <= n; ++i) {
      if (!vis[i]) pri[++cnt] = i, minp[i] = i;
      for (int j = 1; j <= cnt && i * pri[j] <= n; ++j) {
        vis[i * pri[j]] = 1, minp[i * pri[j]] = pri[j];
        if (i % pri[j] == 0) break;
      }
    }
  }
  int n;
  int stk[MAXN], sum[MAXN];
  struct Segment_Tree_II {
    int t[MAXN << 2 | 1];
    Segment_Tree_II() { memset(t, 0x3f, sizeof t); }
    void Chkmin(unsigned ql, unsigned qr, int val, unsigned l = 1,
                unsigned r = n + 1, unsigned num = 1) {
      if (ql > qr) return;
      if (ql <= l && r <= qr) {
        umin(t[num], val);
        return;
      }
      unsigned mid = (l + r) >> 1;
      if (ql <= mid) Chkmin(ql, qr, val, l, mid, num << 1);
      if (qr > mid) Chkmin(ql, qr, val, mid + 1, r, num << 1 | 1);
    }
    void pushall(int* res, unsigned l = 1, unsigned r = n + 1, unsigned num = 1) {
      if (l == r) {
        res[l] = t[num];
        return;
      }
      unsigned mid = (l + r) >> 1;
      umin(t[num << 1], t[num]), umin(t[num << 1 | 1], t[num]);
      pushall(res, l, mid, num << 1), pushall(res, mid + 1, r, num << 1 | 1);
    }
  } T;
  int a[MAXN], f[MAXN];
  char op[MAXN];
  std::vector<std::pair<int, int> > seq[MAXN];
  int main() {
    sieve(1e6);
    n = read();
    for (int i = 1; i <= n; ++i) a[i] = read(), f[i] = n + 1;
    scanf("%s", op + 1);
    for (int i = 1; i <= cnt; ++i)
      seq[pri[i]].emplace_back(std::pair<int, int>(0, 0));
    for (int i = 1; i <= n; ++i) {
      int x = a[i], sgn = (op[i] == '*' ? 1 : -1);
      while (x > 1) {
        int c = 0, fir = minp[x];
        while (minp[x] == fir) x /= fir, ++c;
        seq[fir].emplace_back(std::pair<int, int>(i, c * sgn));
      }
    }
    T.Chkmin(1, n, n);
    for (int w = 1; w <= cnt; ++w) {
      int p = pri[w];
      if (seq[p].size() == 1) continue;
      for (int i = 1; i < seq[p].size(); ++i) {
        sum[i] = sum[i - 1] + seq[p][i].second;
      }
      int top = 0;
      for (int i = int(seq[p].size()) - 1; i >= 0; --i) {
        while (top && sum[stk[top]] >= sum[i]) --top;
        if (top) {
          T.Chkmin(seq[p][i].first + 1, seq[p][stk[top]].first,
                   seq[p][stk[top]].first - 1);
        }
        stk[++top] = i;
      }
    }
    T.pushall(f);
    long long ans = 0;
    for (int i = 1; i <= n; ++i) ans += f[i] - i + 1;
    printf("%lld\n", ans);
    return 0;
  }
  