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;
struct node {
  long long sum, mx;
} tree[4 * 100005];
long long val[100005];
void build(long long pos, long long l, long long r) {
  if (l == r) {
    tree[pos].sum = val[l];
    tree[pos].mx = val[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * pos, l, mid);
  build(2 * pos + 1, mid + 1, r);
  tree[pos].mx = max(tree[2 * pos].mx, tree[2 * pos + 1].mx);
  tree[pos].sum = tree[2 * pos].sum + tree[2 * pos + 1].sum;
}
void updVal(long long pos, long long l, long long r, long long x) {
  if (l == r) {
    tree[pos].sum = val[l];
    tree[pos].mx = val[l];
    return;
  }
  long long mid = (l + r) / 2;
  if (x <= mid)
    updVal(2 * pos, l, mid, x);
  else
    updVal(2 * pos + 1, mid + 1, r, x);
  tree[pos].mx = max(tree[2 * pos].mx, tree[2 * pos + 1].mx);
  tree[pos].sum = tree[2 * pos].sum + tree[2 * pos + 1].sum;
}
long long query(long long pos, long long l, long long r, long long ql,
                long long qr) {
  if (ql > r || l > qr) return 0;
  if (ql <= l && r <= qr) return tree[pos].sum;
  long long mid = (l + r) / 2;
  return query(2 * pos, l, mid, ql, qr) +
         query(2 * pos + 1, mid + 1, r, ql, qr);
}
void updMod(long long pos, long long l, long long r, long long ql, long long qr,
            long long mod) {
  if (ql > r || l > qr || tree[pos].mx < mod) return;
  if (l == r) {
    tree[pos].sum %= mod;
    tree[pos].mx %= mod;
    return;
  }
  long long mid = (l + r) / 2;
  updMod(2 * pos, l, mid, ql, qr, mod);
  updMod(2 * pos + 1, mid + 1, r, ql, qr, mod);
  tree[pos].mx = max(tree[2 * pos].mx, tree[2 * pos + 1].mx);
  tree[pos].sum = tree[2 * pos].sum + tree[2 * pos + 1].sum;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> val[i];
  build(1, 1, n);
  long long a, b, t;
  for (long long i = 1; i <= q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> a >> b;
      cout << query(1, 1, n, a, b) << '\n';
    } else if (t == 2) {
      cin >> a >> b >> t;
      updMod(1, 1, n, a, b, t);
    } else {
      cin >> a >> b;
      val[a] = b;
      updVal(1, 1, n, a);
    }
  }
  return 0;
}