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 M = 1e5 + 5;
long long n, m, a[M];
long long read() {
  register long long x = 0, y = 1;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') y = 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + (c ^ 48), c = getchar();
  }
  return y ? x : -x;
}
struct ed {
  long long l, r, mid, sum, len, lazy, Max;
} tree[M << 2];
void f(long long p, long long sum) {
  tree[p].sum += tree[p].len * sum;
  tree[p].lazy += sum;
}
void push_up(long long p) {
  tree[p].sum = tree[(p) << 1].sum + tree[(p) << 1 | 1].sum;
  tree[p].Max = (tree[(p) << 1].Max) > (tree[(p) << 1 | 1].Max)
                    ? (tree[(p) << 1].Max)
                    : (tree[(p) << 1 | 1].Max);
}
void push_down(long long p) {
  if (tree[p].lazy) {
    f((p) << 1, tree[p].lazy);
    f((p) << 1 | 1, tree[p].lazy);
  }
  tree[p].lazy = 0;
}
void buildtree(long long p, long long l, long long r) {
  tree[p].l = l, tree[p].r = r, tree[p].len = r - l + 1,
  tree[p].mid = (l + r) >> 1;
  if (l == r) {
    tree[p].sum = a[l];
    tree[p].Max = a[l];
    return;
  }
  buildtree((p) << 1, l, tree[p].mid);
  buildtree((p) << 1 | 1, tree[p].mid + 1, r);
  push_up(p);
}
void modify_mod(long long p, long long l, long long r, long long mod) {
  if (tree[p].Max < mod) return;
  if (tree[p].l == tree[p].r) {
    tree[p].sum %= mod;
    tree[p].Max %= mod;
    return;
  }
  if (l <= tree[p].mid) modify_mod((p) << 1, l, r, mod);
  if (r > tree[p].mid) modify_mod((p) << 1 | 1, l, r, mod);
  push_up(p);
}
void modify(long long p, long long x, long long sum) {
  if (tree[p].l == tree[p].r) {
    tree[p].sum = tree[p].Max = sum;
    return;
  }
  if (x <= tree[p].mid) modify((p) << 1, x, sum);
  if (x > tree[p].mid) modify((p) << 1 | 1, x, sum);
  push_up(p);
}
long long query(long long p, long long l, long long r) {
  if (l <= tree[p].l && r >= tree[p].r) return tree[p].sum;
  long long sum = 0;
  push_down(p);
  if (l <= tree[p].mid) sum += query((p) << 1, l, r);
  if (r > tree[p].mid) sum += query((p) << 1 | 1, l, r);
  return sum;
}
signed main() {
  n = read(), m = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  buildtree(1, 1, n);
  for (long long i = 1, l, r, k, x, op, mod; i <= m; i++) {
    op = read();
    if (op == 1) l = read(), r = read(), printf("%lld\n", query(1, l, r));
    if (op == 2) l = read(), r = read(), mod = read(), modify_mod(1, l, r, mod);
    if (op == 3) k = read(), x = read(), modify(1, k, x);
  }
  return 0;
}