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>
const int maxn = 1e5 + 100;
int n, m, a[maxn];
struct node {
  int l, r;
  long long sum, max;
} tree[4 * maxn];
inline void PushUp(int seq) {
  tree[(seq)].sum = tree[(((seq) << 1))].sum + tree[((((seq) << 1) + 1))].sum;
  tree[(seq)].max =
      ((tree[(((seq) << 1))].max) > (tree[((((seq) << 1) + 1))].max)
           ? (tree[(((seq) << 1))].max)
           : (tree[((((seq) << 1) + 1))].max));
}
void Build(int l, int r, int seq) {
  tree[(seq)].l = l, tree[(seq)].r = r;
  if (l == r) {
    tree[(seq)].sum = a[l];
    tree[(seq)].max = a[l];
    return;
  }
  int mid = ((tree[(seq)].l + tree[(seq)].r) >> 1);
  Build(l, mid, ((seq) << 1));
  Build(mid + 1, r, (((seq) << 1) + 1));
  PushUp(seq);
}
void UpdateMod(int l, int r, int seq, int mod) {
  if (tree[(seq)].l == l && tree[(seq)].r == r && tree[(seq)].max < mod) return;
  if (tree[(seq)].l == tree[(seq)].r) {
    tree[(seq)].max %= mod, tree[(seq)].sum %= mod;
    return;
  }
  int mid = ((tree[(seq)].l + tree[(seq)].r) >> 1);
  if (r <= mid)
    UpdateMod(l, r, ((seq) << 1), mod);
  else if (l > mid)
    UpdateMod(l, r, (((seq) << 1) + 1), mod);
  else {
    UpdateMod(l, mid, ((seq) << 1), mod);
    UpdateMod(mid + 1, r, (((seq) << 1) + 1), mod);
  }
  PushUp(seq);
}
void UpdatePoint(int poi, int seq, int val) {
  if (tree[(seq)].l == poi && tree[(seq)].r == poi) {
    tree[(seq)].sum = val, tree[(seq)].max = val;
    return;
  }
  int mid = ((tree[(seq)].l + tree[(seq)].r) >> 1);
  if (poi <= mid)
    UpdatePoint(poi, ((seq) << 1), val);
  else
    UpdatePoint(poi, (((seq) << 1) + 1), val);
  PushUp(seq);
}
long long Query(int l, int r, int seq) {
  if (tree[(seq)].l == l && tree[(seq)].r == r) return tree[(seq)].sum;
  int mid = ((tree[(seq)].l + tree[(seq)].r) >> 1);
  if (r <= mid)
    return Query(l, r, ((seq) << 1));
  else if (l > mid)
    return Query(l, r, (((seq) << 1) + 1));
  else
    return Query(l, mid, ((seq) << 1)) + Query(mid + 1, r, (((seq) << 1) + 1));
}
int main() {
  int ch, l, r, x;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  Build(1, n, 1);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &ch);
    switch (ch) {
      case 1:
        scanf("%d %d", &l, &r);
        printf("%lld\n", Query(l, r, 1));
        break;
      case 2:
        scanf("%d %d %d", &l, &r, &x);
        UpdateMod(l, r, 1, x);
        break;
      case 3:
        scanf("%d %d", &l, &x);
        UpdatePoint(l, 1, x);
        break;
      default:
        break;
    }
  }
  return 0;
}