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 gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
struct segtree {
  struct node {
    long long s, m;
  };
  vector<node> tree;
  long long size;
  void init(long long n) {
    size = 1;
    while (size < n) size *= 2;
    tree.assign(2 * size - 1, {0, 0});
  }
  void build(vector<long long>& arr) {
    init(arr.size());
    build(arr, 0, 0, size);
  }
  void build(vector<long long>& arr, long long x, long long lx, long long rx) {
    if (rx - lx == 1) {
      if (lx < arr.size()) {
        tree[x].s = arr[lx];
        tree[x].m = arr[lx];
      }
      return;
    }
    long long mx = (lx + rx) / 2;
    build(arr, 2 * x + 1, lx, mx);
    build(arr, 2 * x + 2, mx, rx);
    tree[x].s = tree[2 * x + 1].s + tree[2 * x + 2].s;
    tree[x].m = max(tree[2 * x + 1].m, tree[2 * x + 2].m);
  }
  long long sum(long long l, long long r, long long x, long long lx,
                long long rx) {
    if (l <= lx && rx <= r) {
      return tree[x].s;
    }
    if (l >= rx || r <= lx) {
      return 0;
    }
    long long mx = (lx + rx) / 2;
    return sum(l, r, 2 * x + 1, lx, mx) + sum(l, r, 2 * x + 2, mx, rx);
  }
  long long sum(long long l, long long r) { return sum(l, r, 0, 0, size); }
  void propagate(long long x, long long v, long long lx, long long rx) {
    if (tree[x].m < v) return;
    if (rx - lx == 1) {
      tree[x].s %= v;
      tree[x].m = tree[x].s;
      return;
    }
    long long mx = (lx + rx) / 2;
    propagate(2 * x + 1, v, lx, mx);
    propagate(2 * x + 2, v, mx, rx);
    tree[x].s = tree[2 * x + 1].s + tree[2 * x + 2].s;
    tree[x].m = max(tree[2 * x + 1].m, tree[2 * x + 2].m);
  }
  void change(long long l, long long r, long long v, long long x, long long lx,
              long long rx) {
    if (l <= lx && rx <= r) {
      propagate(x, v, lx, rx);
      return;
    }
    if (l >= rx || r <= lx) {
      return;
    }
    long long mx = (lx + rx) / 2;
    change(l, r, v, 2 * x + 1, lx, mx);
    change(l, r, v, 2 * x + 2, mx, rx);
    tree[x].s = tree[2 * x + 1].s + tree[2 * x + 2].s;
    tree[x].m = max(tree[2 * x + 1].m, tree[2 * x + 2].m);
  }
  void change(long long l, long long r, long long v) {
    change(l, r, v, 0, 0, size);
  }
  void add(long long i, long long v, long long x, long long lx, long long rx) {
    if (rx - lx == 1) {
      tree[x].s = v;
      tree[x].m = v;
      return;
    }
    long long mx = (lx + rx) / 2;
    if (i < mx) {
      add(i, v, 2 * x + 1, lx, mx);
    } else {
      add(i, v, 2 * x + 2, mx, rx);
    }
    tree[x].s = tree[2 * x + 1].s + tree[2 * x + 2].s;
    tree[x].m = max(tree[2 * x + 1].m, tree[2 * x + 2].m);
  }
  void add(long long i, long long v) { add(i, v, 0, 0, size); }
};
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  segtree st;
  st.build(arr);
  for (long long i = 0; i < m; i++) {
    int c;
    cin >> c;
    if (c == 1) {
      long long l, r;
      cin >> l >> r;
      l--;
      cout << st.sum(l, r) << '\n';
    } else if (c == 2) {
      long long l, r, v;
      cin >> l >> r >> v;
      l--;
      st.change(l, r, v);
    } else {
      long long k, v;
      cin >> k >> v;
      k--;
      st.add(k, v);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}