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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char ch = 0;
  bool sign = false;
  while (ch < '0' || '9' < ch) sign |= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  x = sign ? -x : x;
}
template <typename T>
inline void print(T x) {
  static char outp[64];
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  int tot = 0;
  while (x) outp[tot++] = x % 10 + '0', x /= 10;
  while (tot) putchar(outp[--tot]);
}
const int MOD = 1e9 + 7;
const int MAXN = 4e5 + 10;
const int MAXT = (1 << 20) + 10;
bool is_p[305];
int ary[MAXN], fmul[305], Set[305];
vector<int> prime;
inline int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = ret * 1ll * x % MOD;
    x = x * 1ll * x % MOD;
    y >>= 1;
  }
  return ret;
}
struct Segment_tree {
  struct Node {
    int l, r;
    int mul, val;
    bitset<64> factor, pd;
  } tree[MAXT];
  bitset<64> qry, prm;
  inline void solve(int x) {
    prm.reset();
    int sq = sqrt(x);
    for (register int i = 1; i <= (int)sq; ++i) {
      if (x % i == 0) {
        if (!is_p[i]) prm[Set[i]] = 1;
        if (!is_p[x / i]) prm[Set[x / i]] = 1;
      }
    }
  }
  inline void update(int root) {
    tree[root].val =
        tree[(root << 1)].val * 1ll * tree[(root << 1 | 1)].val % MOD;
    tree[root].factor |=
        tree[(root << 1)].factor | tree[(root << 1 | 1)].factor;
  }
  inline void pushdown(int root) {
    tree[(root << 1)].mul = tree[(root << 1)].mul * 1ll * tree[root].mul % MOD;
    tree[(root << 1 | 1)].mul =
        tree[(root << 1 | 1)].mul * 1ll * tree[root].mul % MOD;
    tree[(root << 1)].val =
        tree[(root << 1)].val * 1ll *
        qpow(tree[root].mul, tree[(root << 1)].r - tree[(root << 1)].l + 1) %
        MOD;
    tree[(root << 1 | 1)].val =
        tree[(root << 1 | 1)].val * 1ll *
        qpow(tree[root].mul,
             tree[(root << 1 | 1)].r - tree[(root << 1 | 1)].l + 1) %
        MOD;
    tree[(root << 1)].factor |= (tree[(root << 1)].pd |= tree[root].pd);
    tree[(root << 1 | 1)].factor |= (tree[(root << 1 | 1)].pd |= tree[root].pd);
    tree[root].mul = 1;
  }
  inline void buildtree(int root, int l, int r) {
    tree[root].l = l;
    tree[root].r = r;
    tree[root].mul = 1;
    if (l == r) {
      tree[root].val = ary[l];
      int sq = sqrt(ary[l]);
      for (register int i = 1; i <= (int)sq; ++i) {
        if (ary[l] % i == 0) {
          if (!is_p[i]) tree[root].factor[Set[i]] = 1;
          if (!is_p[ary[l] / i]) tree[root].factor[Set[ary[l] / i]] = 1;
        }
      }
      return;
    }
    buildtree((root << 1), l, ((tree[root].l + tree[root].r) >> 1));
    buildtree((root << 1 | 1), ((tree[root].l + tree[root].r) >> 1) + 1, r);
    update(root);
  }
  inline void mul(int root, int l, int r, int k) {
    if (l <= tree[root].l && tree[root].r <= r) {
      tree[root].mul = tree[root].mul * 1ll * k % MOD;
      tree[root].val =
          tree[root].val * 1ll * qpow(k, tree[root].r - tree[root].l + 1) % MOD;
      tree[root].pd |= prm;
      tree[root].factor |= prm;
      return;
    }
    pushdown(root);
    if (l <= ((tree[root].l + tree[root].r) >> 1)) mul((root << 1), l, r, k);
    if (((tree[root].l + tree[root].r) >> 1) < r) mul((root << 1 | 1), l, r, k);
    update(root);
  }
  inline int sum(int root, int l, int r) {
    if (l <= tree[root].l && tree[root].r <= r) {
      qry |= tree[root].factor;
      return tree[root].val;
    }
    pushdown(root);
    int ret = 1;
    if (l <= ((tree[root].l + tree[root].r) >> 1))
      ret = ret * 1ll * sum((root << 1), l, r) % MOD;
    if (((tree[root].l + tree[root].r) >> 1) < r)
      ret = ret * 1ll * sum((root << 1 | 1), l, r) % MOD;
    return ret;
  }
  inline int query(int l, int r) {
    qry.reset();
    long long mm = sum(1, l, r);
    for (register int i = 0; i <= (int)prime.size() - 1; ++i) {
      if (qry[i]) {
        (mm *= (prime[i] - 1) * 1ll * fmul[prime[i]] % MOD) %= MOD;
      }
    }
    return mm;
  }
} Tree;
inline void init() {
  for (register int i = 2; i <= (int)300; ++i) {
    if (!is_p[i]) prime.push_back(i);
    for (auto j : prime) {
      if (i * j > 300) break;
      is_p[i * j] = true;
      if (i % j == 0) break;
    }
  }
  int tot = 0;
  for (auto p : prime) {
    Set[p] = tot;
    ++tot;
  }
  is_p[1] = true;
  for (register int i = 2; i <= (int)300; ++i) {
    fmul[i] = qpow(i, MOD - 2);
  }
}
char s[32];
int main() {
  init();
  int n, m, l, r, x;
  read(n), read(m);
  for (register int i = 1; i <= (int)n; ++i) {
    read(ary[i]);
  }
  Tree.buildtree(1, 1, n);
  while (m--) {
    scanf("%s", s);
    switch (s[0]) {
      case 'T':
        read(l), read(r), print(Tree.query(l, r)), puts("");
        break;
      case 'M':
        read(l), read(r), read(x), Tree.solve(x), Tree.mul(1, l, r, x);
        break;
    }
  }
  return 0;
}
