You are given an array a_1, a_2, …, a_n.

You need to perform q queries of the following two types:

  1. "MULTIPLY l r x" — for every i (l ≤ i ≤ r) multiply a_i by x.
  2. "TOTIENT l r" — print \varphi(∏ _{i=l}^{r} a_i) taken modulo 10^9+7, where \varphi denotes Euler's totient function. 



The [Euler's totient function](http://gg.gg/euler_totient) of a positive integer n (denoted as \varphi(n)) is the number of integers x (1 ≤ x ≤ n) such that \gcd(n,x) = 1.

Input

The first line contains two integers n and q (1 ≤ n ≤ 4 ⋅ 10^5, 1 ≤ q ≤ 2 ⋅ 10^5) — the number of elements in array a and the number of queries.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300) — the elements of array a.

Then q lines follow, describing queries in the format given in the statement.

  1. "MULTIPLY l r x" (1 ≤ l ≤ r ≤ n, 1 ≤ x ≤ 300) — denotes a multiplication query.
  2. "TOTIENT l r" (1 ≤ l ≤ r ≤ n) — denotes a query on the value of Euler's totient function. 



It is guaranteed that there is at least one "TOTIENT" query.

Output

For each "TOTIENT" query, print the answer to it.

Example

Input

4 4
5 9 1 2
TOTIENT 3 3
TOTIENT 3 4
MULTIPLY 4 4 3
TOTIENT 4 4


Output

1
1
2

Note

In the first example, \varphi(1) = 1 for the first query, \varphi(2) = 1 for the second query and \varphi(6) = 2 for the third one.
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const long long int INF = 9e18 + 2e17;
const int inf = 2e9;
const int N = 4e5 + 22;
const double eps = 1e-10;
const double PI = 3.14159265;
int prime[62], prime_inv[62];
int tree[4 * N], lazy_tree[4 * N];
int n, q, L, R, X, a[N];
string s;
long long int mask[4 * N], lazy_mask[4 * N];
bool check[305];
int _mul(int a, int b) { return (1LL * a * b) % mod; }
int _modp(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = _mul(res, a);
    a = _mul(a, a);
    b >>= 1;
  }
  return res;
}
void _prec() {
  for (int i = 2; i * i < 300; i++)
    if (!check[i])
      for (int j = i + i; j < 300; j += i)
        if (!check[j]) check[j] = 1;
  int k = 0;
  for (int i = (int)2; i < (int)300; i++)
    if (!check[i]) prime[k++] = i;
  for (int i = (int)0; i < (int)62; i++)
    prime_inv[i] = _modp(prime[i], mod - 2);
}
long long int _mask(long long int cur_mask, int x) {
  for (int i = (int)0; i < (int)62; i++)
    if (x % prime[i] == 0) cur_mask |= (1LL << i);
  return cur_mask;
}
void _build(int low, int high, int pos) {
  lazy_tree[pos] = 1;
  if (low == high) {
    tree[pos] = a[low];
    mask[pos] = _mask(0, a[low]);
    return;
  }
  int mid = (low + high) / 2;
  _build(low, mid, 2 * pos + 1);
  _build(mid + 1, high, 2 * pos + 2);
  tree[pos] = _mul(tree[2 * pos + 1], tree[2 * pos + 2]);
  mask[pos] = mask[2 * pos + 1] | mask[2 * pos + 2];
}
void _push(int low, int high, int pos) {
  if (lazy_tree[pos] != 1)
    tree[pos] = _mul(tree[pos], _modp(lazy_tree[pos], high - low + 1));
  mask[pos] |= lazy_mask[pos];
  if (low != high) {
    lazy_tree[2 * pos + 1] = _mul(lazy_tree[pos], lazy_tree[2 * pos + 1]);
    lazy_tree[2 * pos + 2] = _mul(lazy_tree[pos], lazy_tree[2 * pos + 2]);
    lazy_mask[2 * pos + 1] |= lazy_mask[pos];
    lazy_mask[2 * pos + 2] |= lazy_mask[pos];
  }
  lazy_tree[pos] = 1;
  lazy_mask[pos] = 0;
}
int _get_mul(int low, int high, int pos) {
  if (L > high || R < low) return 1;
  _push(low, high, pos);
  if (low >= L && high <= R) return tree[pos];
  int mid = (low + high) / 2;
  return _mul(_get_mul(low, mid, 2 * pos + 1),
              _get_mul(mid + 1, high, 2 * pos + 2));
}
void _multiply(int low, int high, int pos) {
  if (L > high || R < low) return;
  _push(low, high, pos);
  if (low >= L && high <= R) {
    lazy_tree[pos] = X;
    lazy_mask[pos] = _mask(0, X);
    _push(low, high, pos);
    return;
  }
  int mid = (low + high) / 2;
  _multiply(low, mid, 2 * pos + 1);
  _multiply(mid + 1, high, 2 * pos + 2);
  tree[pos] = _mul(tree[pos], _modp(X, min(R, high) - max(L, low) + 1));
  mask[pos] = mask[2 * pos + 1] | mask[2 * pos + 2];
}
pair<int, long long int> _totient(int low, int high, int pos) {
  if (L > high || R < low) return {1, 0};
  _push(low, high, pos);
  if (low >= L && high <= R) return {tree[pos], mask[pos]};
  int mid = (low + high) / 2;
  pair<int, long long int> p1, p2;
  p1 = _totient(low, mid, 2 * pos + 1);
  p2 = _totient(mid + 1, high, 2 * pos + 2);
  return {_mul(p1.first, p2.first), p1.second | p2.second};
}
void _solve() {
  _prec();
  cin >> n >> q;
  for (int i = (int)0; i < (int)n; i++) cin >> a[i];
  _build(0, n - 1, 0);
  while (q--) {
    cin >> s >> L >> R;
    L--;
    R--;
    if (s[0] == 'M') {
      cin >> X;
      _multiply(0, n - 1, 0);
    } else {
      pair<int, long long int> p = _totient(0, n - 1, 0);
      int ans = p.first;
      for (int i = (int)0; i < (int)62; i++)
        if ((1LL << i) & p.second)
          ans = _mul(ans, _mul(prime[i] - 1, prime_inv[i]));
      cout << ans << '\n';
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  _solve();
  return 0;
}
