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 + 13;
long long n, m;
struct seg_tree {
  pair<long long, long long> tree[maxN * 4];
  long long get(long long L, long long R, long long l = 0, long long r = n,
                long long indx = 1) {
    if (L <= l && r <= R) return tree[indx].second;
    if (R <= l || r <= L) return 0;
    long long mid = l + r >> 1;
    return get(L, R, l, mid, indx << 1) + get(L, R, mid, r, indx << 1 ^ 1);
  }
  void updbase(long long L, long long R, long long val, long long l = 0,
               long long r = n, long long indx = 1) {
    if (L <= l && r <= R) {
      tree[indx] = {val, val};
      return;
    }
    if (R <= l || r <= L) return;
    long long mid = l + r >> 1;
    updbase(L, R, val, l, mid, indx << 1);
    updbase(L, R, val, mid, r, indx << 1 ^ 1);
    tree[indx] = {max(tree[indx << 1].first, tree[indx << 1 ^ 1].first),
                  tree[indx << 1].second + tree[indx << 1 ^ 1].second};
  }
  void updmod(long long L, long long R, long long mod, long long l = 0,
              long long r = n, long long indx = 1) {
    if (R <= l || r <= L || mod > tree[indx].first) return;
    if (r - l <= 1) {
      tree[indx].second %= mod;
      tree[indx].first %= mod;
      return;
    }
    long long mid = l + r >> 1;
    updmod(L, R, mod, l, mid, indx << 1);
    updmod(L, R, mod, mid, r, indx << 1 ^ 1);
    tree[indx] = {max(tree[indx << 1].first, tree[indx << 1 ^ 1].first),
                  tree[indx << 1].second + tree[indx << 1 ^ 1].second};
  }
} seg;
int32_t main() {
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    seg.updbase(i, i + 1, x);
  }
  while (m--) {
    long long t, l, r, x;
    cin >> t >> l >> r;
    if (t == 1)
      cout << seg.get(l - 1, r) << '\n';
    else if (t == 2) {
      cin >> x;
      seg.updmod(l - 1, r, x);
    } else
      seg.updbase(l - 1, l, r);
  }
}