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>
using namespace std;
const long long mod = 1e9 + 7;
const long long N = 1e6 + 5;
long long n, m, q, a[N];
struct node {
  long long sum, modtag;
  long long left, right;
} tr[N << 2];
inline long long ls(long long pos) { return pos << 1; }
inline long long rs(long long pos) { return pos << 1 | 1; }
inline void push_up(long long pos) {
  tr[pos].modtag = max(tr[ls(pos)].modtag, tr[rs(pos)].modtag);
  tr[pos].sum = tr[ls(pos)].sum + tr[rs(pos)].sum;
  return;
}
void build(long long pos, long long l, long long r) {
  tr[pos].left = l;
  tr[pos].right = r;
  if (l == r) {
    tr[pos].sum = tr[pos].modtag = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(ls(pos), l, mid);
  build(rs(pos), mid + 1, r);
  push_up(pos);
}
void update_mod(long long pos, long long L, long long R, long long sign) {
  if (tr[pos].modtag < sign) return;
  if (tr[pos].left == tr[pos].right) {
    tr[pos].sum %= sign;
    tr[pos].modtag %= sign;
    return;
  }
  long long mid = (tr[pos].left + tr[pos].right) >> 1;
  if (L <= mid) update_mod(ls(pos), L, R, sign);
  if (R > mid) update_mod(rs(pos), L, R, sign);
  push_up(pos);
}
void update_add(long long pos, long long L, long long R, long long sign) {
  if (tr[pos].left == tr[pos].right) {
    tr[pos].sum = tr[pos].modtag = sign;
    return;
  }
  long long mid = (tr[pos].right + tr[pos].left) >> 1;
  if (L <= mid) update_add(ls(pos), L, R, sign);
  if (R > mid) update_add(rs(pos), L, R, sign);
  push_up(pos);
}
long long query(long long pos, long long L, long long R) {
  if (L <= tr[pos].left && tr[pos].right <= R) return tr[pos].sum;
  long long ans = 0, mid = (tr[pos].left + tr[pos].right) >> 1;
  if (L <= mid) ans += query(ls(pos), L, R);
  if (R > mid) ans += query(rs(pos), L, R);
  return ans;
}
inline long long read() {
  long long a = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) a = (a << 3) + (a << 1) + ch - '0';
  return a * f;
}
signed main() {
  n = read(), q = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  build(1, 1, n);
  while (q--) {
    long long opt = read(), x = read(), y = read();
    if (opt == 1)
      printf("%lld\n", query(1, x, y));
    else if (opt == 2)
      update_mod(1, x, y, read());
    else
      update_add(1, x, x, y);
  }
  return 0;
}