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 int maxn = 1e6;
int n, m;
long long a[maxn];
struct seg {
  struct no {
    int l, r;
    long long sum, maxnum;
  } node[maxn << 1];
  inline void up(int now) {
    node[now].sum = node[(now << 1)].sum + node[(now << 1 | 1)].sum;
    node[now].maxnum =
        max(node[(now << 1)].maxnum, node[(now << 1 | 1)].maxnum);
    return;
  }
  void build(int l, int r, int now) {
    node[now].l = l, node[now].r = r;
    if (l == r) return (void)(node[now].sum = node[now].maxnum = a[l]);
    build(l, ((l + r) >> 1), (now << 1)),
        build(((l + r) >> 1) + 1, r, (now << 1 | 1));
    up(now);
    return;
  }
  void chenge(int l, int r, int now, int p) {
    if (node[now].l > r or node[now].r < l) return;
    if (node[now].maxnum < p) return;
    if (node[now].l == node[now].r)
      return (void)(node[now].sum %= p, node[now].maxnum %= p);
    chenge(l, r, (now << 1), p), chenge(l, r, (now << 1 | 1), p);
    up(now);
    return;
  }
  void change(int l, int r, int now, int val) {
    if (node[now].l > r or node[now].r < l) return;
    if (node[now].l == node[now].r)
      return (void)(node[now].sum = node[now].maxnum = val);
    change(l, r, (now << 1), val), change(l, r, (now << 1 | 1), val);
    up(now);
    return;
  }
  void quary(int l, int r, int now, long long &ans) {
    if (node[now].l > r or node[now].r < l) return;
    if (l <= node[now].l and node[now].r <= r)
      return (void)(ans += node[now].sum);
    quary(l, r, (now << 1), ans), quary(l, r, (now << 1 | 1), ans);
    up(now);
    return;
  }
} tree;
long long read_long() {
  long long x = 0, f = 1;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') and (f = -1);
  for (x = ch ^ 48; isdigit(ch = getchar());
       x = (x << 1) + (x << 3) + (ch ^ 48))
    ;
  return x * f;
}
int read_int() {
  int x = 0, f = 1;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') and (f = -1);
  for (x = ch ^ 48; isdigit(ch = getchar());
       x = (x << 1) + (x << 3) + (ch ^ 48))
    ;
  return x * f;
}
signed main() {
  n = read_int(), m = read_int();
  for (int i = 1; i <= n; i++) {
    a[i] = read_long();
  }
  tree.build(1, n, 1);
  for (int opt, x, y, z; m; m--) {
    opt = read_int();
    if (opt == 1) {
      x = read_int(), y = read_int();
      long long ans = 0;
      tree.quary(x, y, 1, ans);
      printf("%lld\n", ans);
    }
    if (opt == 2) {
      x = read_int(), y = read_int(), z = read_int();
      tree.chenge(x, y, 1, z);
    }
    if (opt == 3) {
      x = read_int(), y = read_int();
      tree.change(x, x, 1, y);
    }
  }
  return 0;
}