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;
int a[100005];
struct node {
  int l, r;
  long long val, m;
} node[400005];
inline int read() {
  int x = 0;
  char c;
  c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c <= '9' && c >= '0') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x;
}
inline void up(int u) {
  node[u].m = max(node[u << 1].m, node[u << 1 | 1].m);
  node[u].val = node[u << 1].val + node[u << 1 | 1].val;
}
inline void build(int u, int l, int r) {
  node[u].l = l;
  node[u].r = r;
  if (l == r) {
    node[u].val = node[u].m = a[l];
    return;
  }
  int mid = l + r >> 1;
  build(u << 1, l, mid);
  build(u << 1 | 1, mid + 1, r);
  up(u);
}
inline void mod(int u, int x, int y, int p) {
  if (node[u].m < p) return;
  int l = node[u].l, r = node[u].r;
  if (l == r) {
    node[u].val = node[u].m = node[u].val % p;
    return;
  }
  int mid = l + r >> 1;
  if (mid >= x) mod(u << 1, x, y, p);
  if (mid < y) mod(u << 1 | 1, x, y, p);
  up(u);
}
inline void change(int u, int x, int p) {
  int l = node[u].l, r = node[u].r;
  if (l == r) {
    node[u].val = node[u].m = p;
    return;
  }
  int mid = l + r >> 1;
  if (mid >= x)
    change(u << 1, x, p);
  else
    change(u << 1 | 1, x, p);
  up(u);
}
inline long long find(int u, int x, int y) {
  int l = node[u].l, r = node[u].r;
  if (x <= l && r <= y) return node[u].val;
  int mid = l + r >> 1;
  long long ans = 0;
  if (mid >= x) ans += find(u << 1, x, y);
  if (mid < y) ans += find(u << 1 | 1, x, y);
  return ans;
}
int main() {
  int n, m, x, y, i, p;
  n = read();
  m = read();
  for (i = 1; i <= n; ++i) a[i] = read();
  build(1, 1, n);
  while (m--) {
    i = read();
    x = read();
    y = read();
    if (i == 1)
      printf("%lld\n", find(1, x, y));
    else if (i == 2) {
      p = read();
      mod(1, x, y, p);
    } else
      change(1, x, y);
  }
  return 0;
}