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 maxN = 4e5 + 10;
int n, m, a[maxN];
struct NodeData {
  int L, R, mx;
  long long sum;
};
struct IT {
  NodeData tree[maxN];
  void buildTree(int x, int lef, int rig) {
    tree[x].L = lef;
    tree[x].R = rig;
    if (lef == rig) {
      tree[x].mx = tree[x].sum = a[lef];
      return;
    }
    int mid = (lef + rig) / 2;
    buildTree(x * 2, lef, mid);
    buildTree(x * 2 + 1, mid + 1, rig);
    tree[x] = combine(tree[x * 2], tree[x * 2 + 1]);
  }
  NodeData combine(NodeData lef, NodeData rig) {
    NodeData res;
    res.L = lef.L;
    res.R = rig.R;
    res.mx = max(lef.mx, rig.mx);
    res.sum = lef.sum + rig.sum;
    return res;
  }
  NodeData query1(int x, int lef, int rig) {
    if (lef > tree[x].R || rig < tree[x].L) return tree[0];
    if (lef <= tree[x].L && tree[x].R <= rig) return tree[x];
    return combine(query1(x * 2, lef, rig), query1(x * 2 + 1, lef, rig));
  }
  void query2(int x, int lef, int rig, int val) {
    if (rig < tree[x].L || lef > tree[x].R || tree[x].mx < val) return;
    if (tree[x].L == tree[x].R) {
      tree[x].sum %= val;
      tree[x].mx %= val;
      return;
    }
    query2(x * 2, lef, rig, val);
    query2(x * 2 + 1, lef, rig, val);
    tree[x] = combine(tree[x * 2], tree[x * 2 + 1]);
  }
  void query3(int x, int lef, int rig, int val) {
    if (rig < tree[x].L || lef > tree[x].R) return;
    if (tree[x].L == tree[x].R) {
      tree[x].sum = tree[x].mx = val;
      return;
    }
    query3(x * 2, lef, rig, val);
    query3(x * 2 + 1, lef, rig, val);
    tree[x] = combine(tree[x * 2], tree[x * 2 + 1]);
  }
} Tree;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  Tree.buildTree(1, 1, n);
  for (int i = 1; i <= m; ++i) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      cout << Tree.query1(1, l, r).sum << "\n";
    }
    if (t == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      Tree.query2(1, l, r, x);
    }
    if (t == 3) {
      int k, x;
      cin >> k >> x;
      Tree.query3(1, k, k, x);
    }
  }
}