#include <bits/stdc++.h>
using namespace std;
int read() {
  char ch = getchar();
  int x = 0, pd = 0;
  while (ch < '0' || ch > '9') pd |= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return pd ? -x : x;
}
const int maxn = 1000005;
const int inf = 0x3f3f3f3f;
int n, a[maxn];
char opt[maxn];
int mx, p[maxn], pl, g[maxn], vis[maxn];
void pre() {
  for (int i = 2; i <= mx; i++) {
    if (!vis[i]) p[++pl] = g[i] = i;
    for (int j = 1; j <= pl && i * p[j] <= mx; j++) {
      int x = p[j] * i;
      vis[x] = 1, g[x] = p[j];
      if (i % p[j] == 0) break;
    }
  }
}
int tim[maxn], T;
vector<pair<int, int> > stk[maxn];
int l[maxn];
namespace SGT {
int s[maxn << 2];
void mdf(int o, int l, int r, int x) {
  ++s[o];
  if (l == r) return;
  int mid = l + r >> 1;
  if (x <= mid)
    mdf((o << 1), l, mid, x);
  else
    mdf((o << 1 | 1), mid + 1, r, x);
}
void cover(int o, int l, int r, int L, int R) {
  if ((L <= l && r <= R) || !o[s]) {
    o[s] = 0;
    return;
  }
  int mid = l + r >> 1;
  if (L <= mid) cover((o << 1), l, mid, L, R);
  if (R > mid) cover((o << 1 | 1), mid + 1, r, L, R);
  o[s] = (o << 1)[s] + (o << 1 | 1)[s];
}
}  // namespace SGT
using namespace SGT;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) mx = max(mx, a[i] = read());
  scanf("%s", opt + 1);
  pre();
  for (int i = 1; i <= n; i++)
    if (opt[i] == '*') {
      l[i] = i;
      for (int j = a[i]; j > 1;) {
        int t = g[j], cnt = 0;
        while (g[j] == t) ++cnt, j /= g[j];
        if (tim[t] != T) tim[t] = T, stk[t].clear();
        stk[t].push_back(make_pair(cnt, i));
      }
    } else {
      if (a[i] == 1)
        l[i] = i;
      else
        l[i] = inf;
      for (int j = a[i]; j > 1;) {
        int t = g[j], cnt = 0;
        while (g[j] == t) ++cnt, j /= g[j];
        if (tim[t] != T) tim[t] = T, stk[t].clear();
        while (!stk[t].empty() && cnt > stk[t].back().first)
          cnt -= stk[t].back().first, stk[t].pop_back();
        if (!stk[t].empty()) stk[t].back().first -= cnt, cnt = 0;
        if (cnt) {
          l[i] = 0, ++T;
          break;
        } else
          l[i] = min(l[i], stk[t].back().second);
      }
      if (l[i] == inf) l[i] = 0;
    }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    mdf(1, 1, n, i);
    if (l[i] < i) cover(1, 1, n, l[i] + 1, i);
    ans += s[1];
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int read() {
  char ch = getchar();
  int x = 0, pd = 0;
  while (ch < '0' || ch > '9') pd |= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return pd ? -x : x;
}
const int maxn = 1000005;
const int inf = 0x3f3f3f3f;
int n, a[maxn];
char opt[maxn];
int mx, p[maxn], pl, g[maxn], vis[maxn];
void pre() {
  for (int i = 2; i <= mx; i++) {
    if (!vis[i]) p[++pl] = g[i] = i;
    for (int j = 1; j <= pl && i * p[j] <= mx; j++) {
      int x = p[j] * i;
      vis[x] = 1, g[x] = p[j];
      if (i % p[j] == 0) break;
    }
  }
}
int tim[maxn], T;
vector<pair<int, int> > stk[maxn];
int l[maxn];
namespace SGT {
int s[maxn << 2];
void mdf(int o, int l, int r, int x) {
  ++s[o];
  if (l == r) return;
  int mid = l + r >> 1;
  if (x <= mid)
    mdf((o << 1), l, mid, x);
  else
    mdf((o << 1 | 1), mid + 1, r, x);
}
void cover(int o, int l, int r, int L, int R) {
  if ((L <= l && r <= R) || !o[s]) {
    o[s] = 0;
    return;
  }
  int mid = l + r >> 1;
  if (L <= mid) cover((o << 1), l, mid, L, R);
  if (R > mid) cover((o << 1 | 1), mid + 1, r, L, R);
  o[s] = (o << 1)[s] + (o << 1 | 1)[s];
}
}  // namespace SGT
using namespace SGT;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) mx = max(mx, a[i] = read());
  scanf("%s", opt + 1);
  pre();
  for (int i = 1; i <= n; i++)
    if (opt[i] == '*') {
      l[i] = i;
      for (int j = a[i]; j > 1;) {
        int t = g[j], cnt = 0;
        while (g[j] == t) ++cnt, j /= g[j];
        if (tim[t] != T) tim[t] = T, stk[t].clear();
        stk[t].push_back(make_pair(cnt, i));
      }
    } else {
      if (a[i] == 1)
        l[i] = i;
      else
        l[i] = inf;
      for (int j = a[i]; j > 1;) {
        int t = g[j], cnt = 0;
        while (g[j] == t) ++cnt, j /= g[j];
        if (tim[t] != T) tim[t] = T, stk[t].clear();
        while (!stk[t].empty() && cnt > stk[t].back().first)
          cnt -= stk[t].back().first, stk[t].pop_back();
        if (!stk[t].empty()) stk[t].back().first -= cnt, cnt = 0;
        if (cnt) {
          l[i] = 0, ++T;
          break;
        } else
          l[i] = min(l[i], stk[t].back().second);
      }
      if (l[i] == inf) l[i] = 0;
    }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    mdf(1, 1, n, i);
    if (l[i] < i) cover(1, 1, n, l[i] + 1, i);
    ans += s[1];
  }
  printf("%lld\n", ans);
  return 0;
}

```