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 N = 100010;
int a[N], n, m;
struct Seg {
  int l, r, mx;
  long long sum;
} tree[N * 4];
void BTree(int l, int r, int idx) {
  tree[idx].l = l;
  tree[idx].r = r;
  tree[idx].mx = a[l];
  tree[idx].sum = a[l];
  if (l == r) return;
  int mid = (l + r) >> 1;
  BTree(l, mid, idx << 1);
  BTree(mid + 1, r, idx << 1 | 1);
  tree[idx].mx = max(tree[idx << 1].mx, tree[idx << 1 | 1].mx);
  tree[idx].sum = tree[idx << 1].sum + tree[idx << 1 | 1].sum;
}
void Insert(int l, int r, int x, int idx) {
  if (tree[idx].l == l && tree[idx].r == r) {
    if (tree[idx].mx < x) return;
    if (tree[idx].l != tree[idx].r) {
      int mid = (tree[idx].l + tree[idx].r) >> 1;
      Insert(l, mid, x, idx << 1);
      Insert(mid + 1, r, x, idx << 1 | 1);
      tree[idx].mx = max(tree[idx << 1].mx, tree[idx << 1 | 1].mx);
      tree[idx].sum = tree[idx << 1].sum + tree[idx << 1 | 1].sum;
    } else {
      tree[idx].sum %= x;
      tree[idx].mx %= x;
    }
    return;
  }
  int mid = (tree[idx].l + tree[idx].r) >> 1;
  if (r <= mid)
    Insert(l, r, x, idx << 1);
  else if (l > mid)
    Insert(l, r, x, idx << 1 | 1);
  else
    Insert(l, mid, x, idx << 1), Insert(mid + 1, r, x, idx << 1 | 1);
  tree[idx].mx = max(tree[idx << 1].mx, tree[idx << 1 | 1].mx);
  tree[idx].sum = tree[idx << 1].sum + tree[idx << 1 | 1].sum;
}
void Insert(int id, int x, int idx) {
  if (tree[idx].l == tree[idx].r) {
    tree[idx].mx = x;
    tree[idx].sum = x;
    return;
  }
  int mid = (tree[idx].l + tree[idx].r) >> 1;
  if (id <= mid)
    Insert(id, x, idx << 1);
  else
    Insert(id, x, idx << 1 | 1);
  tree[idx].mx = max(tree[idx << 1].mx, tree[idx << 1 | 1].mx);
  tree[idx].sum = tree[idx << 1].sum + tree[idx << 1 | 1].sum;
}
long long query(int l, int r, int idx) {
  if (tree[idx].l == l && tree[idx].r == r) return tree[idx].sum;
  int mid = (tree[idx].l + tree[idx].r) >> 1;
  long long ret = 0;
  if (r <= mid)
    ret += query(l, r, idx << 1);
  else if (l > mid)
    ret += query(l, r, idx << 1 | 1);
  else
    ret += query(l, mid, idx << 1), ret += query(mid + 1, r, idx << 1 | 1);
  tree[idx].mx = max(tree[idx << 1].mx, tree[idx << 1 | 1].mx);
  tree[idx].sum = tree[idx << 1].sum + tree[idx << 1 | 1].sum;
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  BTree(1, n, 1);
  for (int i = 0; i < m; i++) {
    int t, l, r, k, x;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &l, &r);
      cout << query(l, r, 1) << endl;
    } else if (t == 2) {
      scanf("%d%d%d", &l, &r, &x);
      Insert(l, r, x, 1);
    } else {
      scanf("%d%d", &k, &x);
      Insert(k, x, 1);
    }
  }
  return 0;
}