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;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
bool notPrime[1000005];
long long divPrime[1000005];
int idx[305];
void sieve(int n = 301) {
  int k = sqrt(n);
  memset(notPrime, false, sizeof(notPrime));
  notPrime[1] = true;
  for (int i = 2; i <= k; ++i) {
    if (!notPrime[i]) {
      for (int k = i * i; k <= n; k += i) {
        notPrime[k] = true;
        divPrime[k] = i;
      }
    }
  }
}
vector<long long> pr, inv;
const int N = 400005;
long long a[N];
struct node {
  long long pro;
  bitset<62> bit;
} fake = {1};
node tree[4 * N];
node lazy[4 * N];
inline node combine(node a, node b) {
  node ret;
  ret.pro = a.pro * b.pro % 1000000007;
  ret.bit = a.bit | b.bit;
  return ret;
}
inline node make_node(long long val) {
  node ret;
  ret.pro = val;
  long long k = val;
  ret.bit.reset();
  while (k != 1 && notPrime[k]) {
    long long z = divPrime[k];
    ret.bit[idx[z]] = 1;
    while (k % z == 0) k /= z;
  }
  if (k > 1) ret.bit[idx[k]] = 1;
  return ret;
}
inline void pushdown(long long v, long long st, long long en) {
  if (lazy[v].pro == 1 && lazy[v].bit.count() == 0) return;
  tree[v].pro *= power(lazy[v].pro, en - st + 1, 1000000007) % 1000000007;
  tree[v].pro %= 1000000007;
  tree[v].bit |= lazy[v].bit;
  if (st != en) {
    lazy[v << 1].pro = lazy[v << 1].pro * lazy[v].pro % 1000000007;
    lazy[v << 1 | 1].pro = lazy[v << 1 | 1].pro * lazy[v].pro % 1000000007;
    lazy[v << 1].bit |= lazy[v].bit;
    lazy[v << 1 | 1].bit |= lazy[v].bit;
  }
  lazy[v].pro = 1;
  lazy[v].bit.reset();
  return;
}
void buildTree(long long v, long long st, long long en) {
  if (st == en) {
    tree[v] = make_node(a[st]);
    return;
  }
  long long mid = (st + en) >> 1;
  buildTree(v << 1, st, mid);
  buildTree(v << 1 | 1, mid + 1, en);
  tree[v] = combine(tree[v << 1], tree[v << 1 | 1]);
}
void update(long long v, long long st, long long en, long long in,
            long long val) {
  long long mid = (st + en) >> 1;
  if (st == en) {
    tree[v] = make_node(val);
    a[st] = val;
    return;
  }
  if (in <= mid)
    update(v << 1, st, mid, in, val);
  else
    update(v << 1 | 1, mid + 1, en, in, val);
  tree[v] = combine(tree[v << 1], tree[v << 1 | 1]);
}
void rupdate(long long v, long long st, long long en, long long l, long long r,
             long long val) {
  if (lazy[v].pro != 0) {
    pushdown(v, st, en);
  }
  if (en < l || st > r) return;
  if (st >= l && en <= r) {
    lazy[v].pro = val;
    lazy[v].bit.reset();
    long long k = val;
    while (k != 1 && notPrime[k]) {
      long long z = divPrime[k];
      lazy[v].bit[idx[z]] = 1;
      while (k % z == 0) k /= z;
    }
    if (k > 1) lazy[v].bit[idx[k]] = 1;
    pushdown(v, st, en);
    return;
  }
  long long mid = (st + en) >> 1;
  rupdate(v << 1, st, mid, l, r, val);
  rupdate(v << 1 | 1, mid + 1, en, l, r, val);
  tree[v] = combine(tree[v << 1], tree[v << 1 | 1]);
}
node query(long long v, long long st, long long en, long long l, long long r) {
  if (en < l || st > r) return fake;
  if (lazy[v].pro != 0) {
    pushdown(v, st, en);
  }
  if (st >= l && en <= r) return tree[v];
  long long mid = (st + en) >> 1;
  return combine(query(v << 1, st, mid, l, r),
                 query(v << 1 | 1, mid + 1, en, l, r));
}
long long getans(long long st, long long en, long long l, long long r) {
  node ans = query(1, st, en, l, r);
  long long ret = ans.pro;
  for (long long i = 0; i < 62; ++i) {
    if (ans.bit[i]) {
      ret *= (pr[i] - 1) * inv[i] % 1000000007;
      ret %= 1000000007;
    }
  }
  return ret;
}
int _runtimeTerror_() {
  long long n, i, q;
  cin >> n >> q;
  for (i = 0; i < n; ++i) cin >> a[i];
  buildTree(1, 0, n - 1);
  while (q--) {
    string s;
    long long l, r, val;
    cin >> s >> l >> r;
    if (s == "TOTIENT") {
      cout << getans(0, n - 1, l - 1, r - 1) << "\n";
    } else {
      cin >> val;
      rupdate(1, 0, n - 1, l - 1, r - 1, val);
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sieve();
  for (long long i = 2; i <= 300; ++i)
    if (!notPrime[i])
      pr.push_back(i), inv.push_back(power(i, 1000000007 - 2, 1000000007));
  for (long long i = 0; i < (long long)pr.size(); ++i) idx[pr[i]] = i;
  for (long long i = 0; i < 4 * N; ++i) lazy[i].pro = 1, lazy[i].bit.reset();
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  cerr << "\n" << 1.0 * clock() / CLOCKS_PER_SEC;
  return 0;
}
