#include <bits/stdc++.h>
using namespace std;
template <class T>
void rd(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + ch - '0';
  x *= f;
}
template <class T>
void cmin(T &x, T y) {
  x = min(x, y);
}
template <class T>
void cmax(T &x, T y) {
  x = max(x, y);
}
const long long mod = 1e9 + 7;
template <class T>
T add(T x, T y) {
  return (x + y) % mod;
}
template <class T>
T sub(T x, T y) {
  return (x - y + mod) % mod;
}
template <class T>
T mul(T x, T y) {
  return 1ll * x * y % mod;
}
template <class T>
void Add(T &x, T y) {
  x = (x + y) % mod;
}
template <class T>
void Sub(T &x, T y) {
  x = (x - y + mod) % mod;
}
template <class T>
void Mul(T &x, T y) {
  x = 1ll * x * y % mod;
}
template <class T>
T qpow(T a, T b = mod - 2) {
  T ret = 1;
  for (; b; b >>= 1, Mul(a, a))
    if (b & 1) Mul(ret, a);
  return ret;
}
const long long N = 1e6 + 10, V = 1e6;
long long n, a[N], L[N], top;
pair<long long, long long> stk[N];
vector<long long> vec[V + 5];
char s[N];
long long pri[V + 5], mark[V + 5], cnt, mn[V + 5];
void sieve(long long n) {
  for (long long i = 2; i <= n; i++) {
    if (!mark[i]) pri[++cnt] = i, mn[i] = i;
    for (long long j = 1; j <= cnt && i * pri[j] <= n; j++) {
      mark[i * pri[j]] = 1, mn[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
void ins(long long x, long long id) {
  L[id] = id;
  while (x > 1) {
    vec[mn[x]].push_back(id), x /= mn[x];
  }
}
void del(long long x, long long id) {
  if (x == 1) {
    L[id] = id;
    return;
  }
  L[id] = id;
  while (x > 1) {
    if (vec[mn[x]].empty()) {
      L[id] = 0;
      return;
    }
    cmin(L[id], vec[mn[x]].back());
    vec[mn[x]].pop_back();
    x /= mn[x];
  }
}
signed main() {
  sieve(V);
  rd(n);
  for (long long i = 1; i <= n; i++) rd(a[i]);
  scanf("%s", s + 1);
  for (long long i = 1; i <= n; i++) {
    if (s[i] == '*')
      ins(a[i], i);
    else
      del(a[i], i);
  }
  long long pos = 0;
  long long ans = 0;
  for (long long l = n; l >= 1; l--) {
    long long cnt = 1;
    while (top && stk[top].first >= L[l]) {
      if (pos == top) --pos;
      cnt += stk[top].second;
      --top;
    }
    stk[++top] = make_pair(L[l], cnt);
    if (L[l] == l) ans += cnt;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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;
template <class T>
void rd(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + ch - '0';
  x *= f;
}
template <class T>
void cmin(T &x, T y) {
  x = min(x, y);
}
template <class T>
void cmax(T &x, T y) {
  x = max(x, y);
}
const long long mod = 1e9 + 7;
template <class T>
T add(T x, T y) {
  return (x + y) % mod;
}
template <class T>
T sub(T x, T y) {
  return (x - y + mod) % mod;
}
template <class T>
T mul(T x, T y) {
  return 1ll * x * y % mod;
}
template <class T>
void Add(T &x, T y) {
  x = (x + y) % mod;
}
template <class T>
void Sub(T &x, T y) {
  x = (x - y + mod) % mod;
}
template <class T>
void Mul(T &x, T y) {
  x = 1ll * x * y % mod;
}
template <class T>
T qpow(T a, T b = mod - 2) {
  T ret = 1;
  for (; b; b >>= 1, Mul(a, a))
    if (b & 1) Mul(ret, a);
  return ret;
}
const long long N = 1e6 + 10, V = 1e6;
long long n, a[N], L[N], top;
pair<long long, long long> stk[N];
vector<long long> vec[V + 5];
char s[N];
long long pri[V + 5], mark[V + 5], cnt, mn[V + 5];
void sieve(long long n) {
  for (long long i = 2; i <= n; i++) {
    if (!mark[i]) pri[++cnt] = i, mn[i] = i;
    for (long long j = 1; j <= cnt && i * pri[j] <= n; j++) {
      mark[i * pri[j]] = 1, mn[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
void ins(long long x, long long id) {
  L[id] = id;
  while (x > 1) {
    vec[mn[x]].push_back(id), x /= mn[x];
  }
}
void del(long long x, long long id) {
  if (x == 1) {
    L[id] = id;
    return;
  }
  L[id] = id;
  while (x > 1) {
    if (vec[mn[x]].empty()) {
      L[id] = 0;
      return;
    }
    cmin(L[id], vec[mn[x]].back());
    vec[mn[x]].pop_back();
    x /= mn[x];
  }
}
signed main() {
  sieve(V);
  rd(n);
  for (long long i = 1; i <= n; i++) rd(a[i]);
  scanf("%s", s + 1);
  for (long long i = 1; i <= n; i++) {
    if (s[i] == '*')
      ins(a[i], i);
    else
      del(a[i], i);
  }
  long long pos = 0;
  long long ans = 0;
  for (long long l = n; l >= 1; l--) {
    long long cnt = 1;
    while (top && stk[top].first >= L[l]) {
      if (pos == top) --pos;
      cnt += stk[top].second;
      --top;
    }
    stk[++top] = make_pair(L[l], cnt);
    if (L[l] == l) ans += cnt;
  }
  printf("%lld\n", ans);
  return 0;
}

```