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 INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int T = 1e5 + 5;
long long v[T];
int n, m, a, b, op;
long long val;
long long seg[4 * T];
long long lazy[4 * T];
int ass[4 * T];
long long small[4 * T];
long long big[4 * T];
void build(int node, int i, int j) {
  if (i == j)
    seg[node] = big[node] = small[node] = v[i];
  else {
    int mid = (i + j) >> 1;
    build(2 * node, i, mid);
    build(2 * node + 1, mid + 1, j);
    seg[node] = seg[2 * node] + seg[2 * node + 1];
    big[node] = max(big[2 * node], big[2 * node + 1]);
    small[node] = min(small[2 * node], small[2 * node + 1]);
  }
}
void prop(int node, int i, int j) {
  if (!ass[node]) return;
  small[node] = big[node] = lazy[node];
  seg[node] = (long long)(j - i + 1) * lazy[node];
  if (i != j) {
    lazy[2 * node] = lazy[2 * node + 1] = lazy[node];
    ass[2 * node] = ass[2 * node + 1] = 1;
  }
  lazy[node] = 0;
  ass[node] = 0;
}
void update(int node, int i, int j) {
  prop(node, i, j);
  if (i > b or j < a or (op == 2 and big[node] < val))
    return;
  else if (i >= a and j <= b and big[node] == small[node]) {
    if (op == 2)
      lazy[node] = big[node] % val;
    else
      lazy[node] = val;
    ass[node] = 1;
    prop(node, i, j);
  } else {
    int mid = (i + j) >> 1;
    update(2 * node, i, mid);
    update(2 * node + 1, mid + 1, j);
    seg[node] = seg[2 * node] + seg[2 * node + 1];
    big[node] = max(big[2 * node], big[2 * node + 1]);
    small[node] = min(small[2 * node], small[2 * node + 1]);
  }
}
long long query(int node, int i, int j) {
  prop(node, i, j);
  if (i > b or j < a) return 0;
  if (i >= a and j <= b) return seg[node];
  int mid = (i + j) >> 1;
  return query(2 * node, i, mid) + query(2 * node + 1, mid + 1, j);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> v[i];
  build(1, 1, n);
  while (m--) {
    cin >> op >> a >> b;
    if (op == 1)
      cout << query(1, 1, n) << '\n';
    else if (op == 2)
      cin >> val, update(1, 1, n);
    else
      val = b, b = a, update(1, 1, n);
  }
  return 0;
}