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 MAXN = 1e5 + 10;
long long n, m, rval[MAXN];
struct Node {
  long long maxn, sum;
} nodes[MAXN * 4];
void pushup(long long x) {
  (nodes[x]).maxn = max(nodes[(x << 1)].maxn, nodes[(x << 1 | 1)].maxn);
  (nodes[x]).sum = nodes[(x << 1)].sum + nodes[(x << 1 | 1)].sum;
}
void build(long long x, long long l, long long r) {
  if (l == r)
    (nodes[x]).maxn = (nodes[x]).sum = rval[l];
  else
    build((x << 1), l, ((l + r) >> 1)),
        build((x << 1 | 1), (((l + r) >> 1) + 1), r), pushup(x);
}
void update_range(long long x, long long l, long long r, long long tl,
                  long long tr, long long val) {
  if ((nodes[x]).maxn < val) return;
  if (l == r) {
    (nodes[x]).sum = (nodes[x]).sum % val,
    (nodes[x]).maxn = (nodes[x]).maxn % val;
    return;
  }
  if (tl <= ((l + r) >> 1))
    update_range((x << 1), l, ((l + r) >> 1), tl, tr, val);
  if (tr >= (((l + r) >> 1) + 1))
    update_range((x << 1 | 1), (((l + r) >> 1) + 1), r, tl, tr, val);
  pushup(x);
}
void update_one(long long x, long long l, long long r, long long t,
                long long val) {
  if (l == r) {
    (nodes[x]).sum = (nodes[x]).maxn = val;
    return;
  }
  if (t <= ((l + r) >> 1))
    update_one((x << 1), l, ((l + r) >> 1), t, val);
  else
    update_one((x << 1 | 1), (((l + r) >> 1) + 1), r, t, val);
  pushup(x);
}
long long query(long long x, long long l, long long r, long long tl,
                long long tr) {
  if (tl <= l && r <= tr) return (nodes[x]).sum;
  long long res = 0;
  if (tl <= ((l + r) >> 1)) res += query((x << 1), l, ((l + r) >> 1), tl, tr);
  if (tr >= (((l + r) >> 1) + 1))
    res += query((x << 1 | 1), (((l + r) >> 1) + 1), r, tl, tr);
  return res;
}
signed main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld", rval + i);
  build(1, 1, n);
  for (long long i = 1, op, x, y, z; i <= m; i++) {
    scanf("%lld", &op);
    switch (op) {
      case 1:
        scanf("%lld%lld", &x, &y), printf("%lld\n", query(1, 1, n, x, y));
        break;
      case 2:
        scanf("%lld%lld%lld", &x, &y, &z), update_range(1, 1, n, x, y, z);
        break;
      case 3:
        scanf("%lld%lld", &x, &y), update_one(1, 1, n, x, y);
    }
  }
  return 0;
}