#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long _INF = 0xc0c0c0c0c0c0c0c0;
const long long mod = (int)1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long ksm(long long a, long long b, long long mod) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long inv2(long long a, long long mod) { return ksm(a, mod - 2, mod); }
const int MAX_N = 4e5 + 5, M = 301, Lim = 62;
int P[Lim] = {2,   3,   5,   7,   11,  13,  17,  19,  23,  29,  31,  37,  41,
              43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89,  97,  101,
              103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167,
              173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239,
              241, 251, 257, 263, 269, 271, 277, 281, 283, 293};
int rev[M] = {
    0, 0,  0, 1,  0, 2,  0, 3,  0, 0,  0, 4, 0, 5,  0, 0,  0, 6,  0, 7,  0, 0,
    0, 8,  0, 0,  0, 0,  0, 9,  0, 10, 0, 0, 0, 0,  0, 11, 0, 0,  0, 12, 0, 13,
    0, 0,  0, 14, 0, 0,  0, 0,  0, 15, 0, 0, 0, 0,  0, 16, 0, 17, 0, 0,  0, 0,
    0, 18, 0, 0,  0, 19, 0, 20, 0, 0,  0, 0, 0, 21, 0, 0,  0, 22, 0, 0,  0, 0,
    0, 23, 0, 0,  0, 0,  0, 0,  0, 24, 0, 0, 0, 25, 0, 26, 0, 0,  0, 27, 0, 28,
    0, 0,  0, 29, 0, 0,  0, 0,  0, 0,  0, 0, 0, 0,  0, 0,  0, 30, 0, 0,  0, 31,
    0, 0,  0, 0,  0, 32, 0, 33, 0, 0,  0, 0, 0, 0,  0, 0,  0, 34, 0, 35, 0, 0,
    0, 0,  0, 36, 0, 0,  0, 0,  0, 37, 0, 0, 0, 38, 0, 0,  0, 0,  0, 39, 0, 0,
    0, 0,  0, 40, 0, 41, 0, 0,  0, 0,  0, 0, 0, 0,  0, 42, 0, 43, 0, 0,  0, 44,
    0, 45, 0, 0,  0, 0,  0, 0,  0, 0,  0, 0, 0, 46, 0, 0,  0, 0,  0, 0,  0, 0,
    0, 0,  0, 47, 0, 0,  0, 48, 0, 49, 0, 0, 0, 50, 0, 0,  0, 0,  0, 51, 0, 52,
    0, 0,  0, 0,  0, 0,  0, 0,  0, 53, 0, 0, 0, 0,  0, 54, 0, 0,  0, 0,  0, 55,
    0, 0,  0, 0,  0, 56, 0, 57, 0, 0,  0, 0, 0, 58, 0, 0,  0, 59, 0, 60, 0, 0,
    0, 0,  0, 0,  0, 0,  0, 61, 0, 0,  0, 0, 0, 0,  0};
long long val[MAX_N << 2], col[MAX_N << 2];
long long now, Ans, s[MAX_N << 2], colf[MAX_N << 2];
void up(int rt, int l, int r) {
  s[rt] = s[rt << 1] | s[rt << 1 | 1];
  val[rt] = 1ll * val[rt << 1] * val[rt << 1 | 1] % 1000000007;
}
long long FP(int x, int k) {
  long long t = 1;
  for (; k; k >>= 1, x = 1ll * x * x % 1000000007)
    if (k & 1) t = 1ll * t * x % 1000000007;
  return t;
}
void update(int rt, int l, int r, long long v, long long vf) {
  val[rt] = 1ll * val[rt] * FP(v, r - l + 1) % 1000000007;
  col[rt] = 1ll * col[rt] * v % 1000000007;
  s[rt] |= vf;
  colf[rt] |= vf;
}
void build(int rt, int l, int r) {
  col[rt] = 1;
  if (l == r) {
    int v;
    scanf("%d", &v);
    val[rt] = v;
    for (int i = 0; i < Lim && P[i] * P[i] <= v; ++i) {
      if (v % P[i] == 0) {
        s[rt] |= 1ll << i, v /= P[i];
        while (v % P[i] == 0) v /= P[i];
      }
    }
    if (v != 1) s[rt] |= 1ll << rev[v];
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid), build(rt << 1 | 1, mid + 1, r);
  up(rt, l, r);
}
void down(int rt, int l, int r) {
  int mid = (l + r) >> 1;
  update(rt << 1, l, mid, col[rt], colf[rt]);
  update(rt << 1 | 1, mid + 1, r, col[rt], colf[rt]);
  col[rt] = 1;
  colf[rt] = 0;
}
void modify(int rt, int l, int r, int x, int y, long long v) {
  if (x <= l && r <= y) {
    update(rt, l, r, v, now);
    return;
  }
  if (col[rt] != 1) down(rt, l, r);
  int mid = (l + r) >> 1;
  if (x <= mid) modify(rt << 1, l, mid, x, y, v);
  if (mid < y) modify(rt << 1 | 1, mid + 1, r, x, y, v);
  up(rt, l, r);
}
long long query(int rt, int l, int r, int x, int y) {
  if (x <= l && r <= y) return Ans |= s[rt], val[rt];
  if (col[rt] != 1) down(rt, l, r);
  int mid = (l + r) >> 1;
  if (x <= mid)
    if (mid < y)
      return 1ll * query(rt << 1, l, mid, x, y) *
             query(rt << 1 | 1, mid + 1, r, x, y) % 1000000007;
    else
      return query(rt << 1, l, mid, x, y);
  return query(rt << 1 | 1, mid + 1, r, x, y);
}
int inv[M], coef[M];
char str[15];
int main() {
  inv[1] = 1;
  for (int i = 2; i < M; ++i)
    inv[i] =
        1ll * (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  for (int i = 0; i < Lim; i++)
    coef[i] = 1ll * (P[i] - 1) * inv[P[i]] % 1000000007;
  int n, q, x, y;
  scanf("%d%d", &n, &q);
  build(1, 1, n);
  while (q--) {
    scanf("%s", str);
    if (str[0] == 'T') {
      Ans = 0;
      scanf("%d%d", &x, &y);
      long long VAL = query(1, 1, n, x, y);
      for (int i = 0; i < Lim; ++i)
        if (Ans >> i & 1) VAL = 1ll * VAL * coef[i] % 1000000007;
      printf("%lld\n", VAL);
    } else {
      int V, tmp;
      scanf("%d%d%d", &x, &y, &tmp);
      V = tmp;
      now = 0;
      for (int i = 0; i < Lim && P[i] * P[i] <= V; ++i) {
        if (V % P[i] == 0) {
          now |= 1 << i, V /= P[i];
          while (V % P[i] == 0) V /= P[i];
        }
      }
      if (V != 1) now |= 1ll << rev[V];
      modify(1, 1, n, x, y, tmp);
    }
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long _INF = 0xc0c0c0c0c0c0c0c0;
const long long mod = (int)1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long ksm(long long a, long long b, long long mod) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long inv2(long long a, long long mod) { return ksm(a, mod - 2, mod); }
const int MAX_N = 4e5 + 5, M = 301, Lim = 62;
int P[Lim] = {2,   3,   5,   7,   11,  13,  17,  19,  23,  29,  31,  37,  41,
              43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89,  97,  101,
              103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167,
              173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239,
              241, 251, 257, 263, 269, 271, 277, 281, 283, 293};
int rev[M] = {
    0, 0,  0, 1,  0, 2,  0, 3,  0, 0,  0, 4, 0, 5,  0, 0,  0, 6,  0, 7,  0, 0,
    0, 8,  0, 0,  0, 0,  0, 9,  0, 10, 0, 0, 0, 0,  0, 11, 0, 0,  0, 12, 0, 13,
    0, 0,  0, 14, 0, 0,  0, 0,  0, 15, 0, 0, 0, 0,  0, 16, 0, 17, 0, 0,  0, 0,
    0, 18, 0, 0,  0, 19, 0, 20, 0, 0,  0, 0, 0, 21, 0, 0,  0, 22, 0, 0,  0, 0,
    0, 23, 0, 0,  0, 0,  0, 0,  0, 24, 0, 0, 0, 25, 0, 26, 0, 0,  0, 27, 0, 28,
    0, 0,  0, 29, 0, 0,  0, 0,  0, 0,  0, 0, 0, 0,  0, 0,  0, 30, 0, 0,  0, 31,
    0, 0,  0, 0,  0, 32, 0, 33, 0, 0,  0, 0, 0, 0,  0, 0,  0, 34, 0, 35, 0, 0,
    0, 0,  0, 36, 0, 0,  0, 0,  0, 37, 0, 0, 0, 38, 0, 0,  0, 0,  0, 39, 0, 0,
    0, 0,  0, 40, 0, 41, 0, 0,  0, 0,  0, 0, 0, 0,  0, 42, 0, 43, 0, 0,  0, 44,
    0, 45, 0, 0,  0, 0,  0, 0,  0, 0,  0, 0, 0, 46, 0, 0,  0, 0,  0, 0,  0, 0,
    0, 0,  0, 47, 0, 0,  0, 48, 0, 49, 0, 0, 0, 50, 0, 0,  0, 0,  0, 51, 0, 52,
    0, 0,  0, 0,  0, 0,  0, 0,  0, 53, 0, 0, 0, 0,  0, 54, 0, 0,  0, 0,  0, 55,
    0, 0,  0, 0,  0, 56, 0, 57, 0, 0,  0, 0, 0, 58, 0, 0,  0, 59, 0, 60, 0, 0,
    0, 0,  0, 0,  0, 0,  0, 61, 0, 0,  0, 0, 0, 0,  0};
long long val[MAX_N << 2], col[MAX_N << 2];
long long now, Ans, s[MAX_N << 2], colf[MAX_N << 2];
void up(int rt, int l, int r) {
  s[rt] = s[rt << 1] | s[rt << 1 | 1];
  val[rt] = 1ll * val[rt << 1] * val[rt << 1 | 1] % 1000000007;
}
long long FP(int x, int k) {
  long long t = 1;
  for (; k; k >>= 1, x = 1ll * x * x % 1000000007)
    if (k & 1) t = 1ll * t * x % 1000000007;
  return t;
}
void update(int rt, int l, int r, long long v, long long vf) {
  val[rt] = 1ll * val[rt] * FP(v, r - l + 1) % 1000000007;
  col[rt] = 1ll * col[rt] * v % 1000000007;
  s[rt] |= vf;
  colf[rt] |= vf;
}
void build(int rt, int l, int r) {
  col[rt] = 1;
  if (l == r) {
    int v;
    scanf("%d", &v);
    val[rt] = v;
    for (int i = 0; i < Lim && P[i] * P[i] <= v; ++i) {
      if (v % P[i] == 0) {
        s[rt] |= 1ll << i, v /= P[i];
        while (v % P[i] == 0) v /= P[i];
      }
    }
    if (v != 1) s[rt] |= 1ll << rev[v];
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid), build(rt << 1 | 1, mid + 1, r);
  up(rt, l, r);
}
void down(int rt, int l, int r) {
  int mid = (l + r) >> 1;
  update(rt << 1, l, mid, col[rt], colf[rt]);
  update(rt << 1 | 1, mid + 1, r, col[rt], colf[rt]);
  col[rt] = 1;
  colf[rt] = 0;
}
void modify(int rt, int l, int r, int x, int y, long long v) {
  if (x <= l && r <= y) {
    update(rt, l, r, v, now);
    return;
  }
  if (col[rt] != 1) down(rt, l, r);
  int mid = (l + r) >> 1;
  if (x <= mid) modify(rt << 1, l, mid, x, y, v);
  if (mid < y) modify(rt << 1 | 1, mid + 1, r, x, y, v);
  up(rt, l, r);
}
long long query(int rt, int l, int r, int x, int y) {
  if (x <= l && r <= y) return Ans |= s[rt], val[rt];
  if (col[rt] != 1) down(rt, l, r);
  int mid = (l + r) >> 1;
  if (x <= mid)
    if (mid < y)
      return 1ll * query(rt << 1, l, mid, x, y) *
             query(rt << 1 | 1, mid + 1, r, x, y) % 1000000007;
    else
      return query(rt << 1, l, mid, x, y);
  return query(rt << 1 | 1, mid + 1, r, x, y);
}
int inv[M], coef[M];
char str[15];
int main() {
  inv[1] = 1;
  for (int i = 2; i < M; ++i)
    inv[i] =
        1ll * (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  for (int i = 0; i < Lim; i++)
    coef[i] = 1ll * (P[i] - 1) * inv[P[i]] % 1000000007;
  int n, q, x, y;
  scanf("%d%d", &n, &q);
  build(1, 1, n);
  while (q--) {
    scanf("%s", str);
    if (str[0] == 'T') {
      Ans = 0;
      scanf("%d%d", &x, &y);
      long long VAL = query(1, 1, n, x, y);
      for (int i = 0; i < Lim; ++i)
        if (Ans >> i & 1) VAL = 1ll * VAL * coef[i] % 1000000007;
      printf("%lld\n", VAL);
    } else {
      int V, tmp;
      scanf("%d%d%d", &x, &y, &tmp);
      V = tmp;
      now = 0;
      for (int i = 0; i < Lim && P[i] * P[i] <= V; ++i) {
        if (V % P[i] == 0) {
          now |= 1 << i, V /= P[i];
          while (V % P[i] == 0) V /= P[i];
        }
      }
      if (V != 1) now |= 1ll << rev[V];
      modify(1, 1, n, x, y, tmp);
    }
  }
  return 0;
}

```