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;
long long _max[100010 << 2], sum[100010 << 2];
void push_up(int k) {
  sum[k] = sum[k << 1] + sum[k << 1 | 1];
  _max[k] = max(_max[k << 1], _max[k << 1 | 1]);
}
void build(int k, int l, int r) {
  _max[k] = sum[k] = 0;
  if (l == r) {
    cin >> sum[k];
    _max[k] = sum[k];
    return;
  }
  int mid = (l + r) >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  push_up(k);
}
long long query(int k, int l, int r, int a, int b) {
  if (a <= l && r <= b) return sum[k];
  int mid = (l + r) >> 1;
  long long ret = 0;
  if (a <= mid) ret += query(k << 1, l, mid, a, b);
  if (b > mid) ret += query(k << 1 | 1, mid + 1, r, a, b);
  return ret;
}
void update(int k, int l, int r, int x, int y) {
  if (l == r) {
    sum[k] = _max[k] = y;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    update(k << 1, l, mid, x, y);
  else
    update(k << 1 | 1, mid + 1, r, x, y);
  push_up(k);
}
void mood(int k, int l, int r, int a, int b, int m) {
  if (_max[k] < m) return;
  if (l == r) {
    sum[k] = sum[k] % m;
    _max[k] = sum[k];
    return;
  }
  int mid = (l + r) >> 1;
  if (a <= mid) mood(k << 1, l, mid, a, b, m);
  if (b > mid) mood(k << 1 | 1, mid + 1, r, a, b, m);
  push_up(k);
}
int main() {
  int n, m, q, op, l, r;
  scanf("%d%d", &n, &q);
  build(1, 1, n);
  while (q--) {
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1) {
      cout << query(1, 1, n, l, r) << endl;
    } else if (op == 3) {
      update(1, 1, n, l, r);
    } else {
      scanf("%d", &m);
      mood(1, 1, n, l, r, m);
    }
  }
}