Problem: At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

  1. Print operation l, r. Picks should write down the value of <image>. 
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 
  3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x). 



Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.

Each of the next m lines begins with a number type <image>. 

  * If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1. 
  * If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2. 
  * If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3. 

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3


Output

8
5


Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10


Output

49
15
23
1
9

Note

Consider the first testcase:

  * At first, a = {1, 2, 3, 4, 5}. 
  * After operation 1, a = {1, 2, 3, 0, 1}. 
  * After operation 2, a = {1, 2, 5, 0, 1}. 
  * At operation 3, 2 + 5 + 0 + 1 = 8. 
  * After operation 4, a = {1, 2, 2, 0, 1}. 
  * At operation 5, 1 + 2 + 2 = 5. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
namespace _ {
namespace IO_IN {
char buf[1 << 23], *p1 = buf, *p2 = buf;
int getc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 23, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <class T>
inline T read() {
  char ch;
  bool flag = 0;
  T x = 0;
  while (ch = getc(), !isdigit(ch))
    if (ch == '-') flag = 1;
  while (isdigit(ch)) x = x * 10 + ch - 48, ch = getc();
  return flag ? -x : x;
}
struct {
  inline operator int() { return read<int>(); }
  inline operator long long() { return read<long long>(); }
  template <class T>
  inline void operator()(T& x) {
    x = *this;
  }
  template <class T, class... A>
  inline void operator()(T& x, A&... a) {
    x = *this, this->operator()(a...);
  }
} IN;
}  // namespace IO_IN
using namespace IO_IN;
namespace Std_Function {
template <class T>
inline T Max(T x, T y) {
  return x > y ? x : y;
}
template <class T>
inline T Min(T x, T y) {
  return x < y ? x : y;
}
template <class T>
inline bool chkmax(T& x, T y) {
  x = x > y ? x : y;
  return x == y;
}
template <class T>
inline bool chkmin(T& x, T y) {
  x = x < y ? x : y;
  return x == y;
}
}  // namespace Std_Function
using namespace Std_Function;
}  // namespace _
using namespace _;
const int N = 1e5 + 5;
int n, m, a[N];
struct Segment_Tree {
  int max[N << 2];
  long long sum[N << 2];
  inline void pushup(int x) {
    sum[x] = sum[((x) << 1)] + sum[((x) << 1 | 1)],
    max[x] = Max(max[((x) << 1)], max[((x) << 1 | 1)]);
  }
  void build(int x, int l, int r) {
    if (l == r) return sum[x] = max[x] = a[l], void();
    build(((x) << 1), l, ((l + r) >> 1)),
        build(((x) << 1 | 1), ((l + r) >> 1) + 1, r), pushup(x);
  }
  void mod(int x, int l, int r, int L, int R, int p) {
    if (max[x] < p) return;
    if (l == r) return max[x] = sum[x] = sum[x] % p, void();
    if (L <= ((l + r) >> 1)) mod(((x) << 1), l, ((l + r) >> 1), L, R, p);
    if (R > ((l + r) >> 1)) mod(((x) << 1 | 1), ((l + r) >> 1) + 1, r, L, R, p);
    pushup(x);
  }
  void change(int x, int l, int r, int pos, int val) {
    if (l == r) return max[x] = sum[x] = val, void();
    pos <= ((l + r) >> 1)
        ? change(((x) << 1), l, ((l + r) >> 1), pos, val)
        : change(((x) << 1 | 1), ((l + r) >> 1) + 1, r, pos, val);
    pushup(x);
  }
  long long query(int x, int l, int r, int L, int R) {
    if (L == l && r == R) return sum[x];
    if (R <= ((l + r) >> 1))
      return query(((x) << 1), l, ((l + r) >> 1), L, R);
    else if (L > ((l + r) >> 1))
      return query(((x) << 1 | 1), ((l + r) >> 1) + 1, r, L, R);
    else
      return query(((x) << 1), l, ((l + r) >> 1), L, ((l + r) >> 1)) +
             query(((x) << 1 | 1), ((l + r) >> 1) + 1, r, ((l + r) >> 1) + 1,
                   R);
  }
} Tree;
int op, l, r, k, x;
int main() {
  IN(n, m);
  for (int i = 1; i <= n; ++i) IN(a[i]);
  Tree.build(1, 1, n);
  while (m--) {
    IN(op);
    if (op == 1) IN(l, r), printf("%lld\n", Tree.query(1, 1, n, l, r));
    if (op == 2) IN(l, r, x), Tree.mod(1, 1, n, l, r, x);
    if (op == 3) IN(k, x), Tree.change(1, 1, n, k, x);
  }
  return 0;
}