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 = 1e5 + 5;
int a[maxn];
struct seg {
  int l, r;
  long long sum, v;
} t[maxn << 2];
void build(int o, int l, int r) {
  t[o].l = l, t[o].r = r;
  if (l == r) {
    t[o].sum = a[l];
    t[o].v = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
  t[o].sum = t[o << 1].sum + t[o << 1 | 1].sum;
  t[o].v = max(t[o << 1].v, t[o << 1 | 1].v);
}
void update1(int o, int l, int r, long long d) {
  if (t[o].v < d) return;
  if (t[o].l == t[o].r) {
    t[o].sum %= d;
    t[o].v %= d;
    return;
  }
  int mid = (t[o].l + t[o].r) >> 1;
  if (l <= mid) update1(o << 1, l, r, d);
  if (r > mid) update1(o << 1 | 1, l, r, d);
  t[o].sum = t[o << 1].sum + t[o << 1 | 1].sum;
  t[o].v = max(t[o << 1].v, t[o << 1 | 1].v);
}
void update2(int o, int l, int r, long long p) {
  if (l <= t[o].l && r >= t[o].r) {
    t[o].sum = p;
    t[o].v = p;
    return;
  }
  int mid = (t[o].l + t[o].r) >> 1;
  if (l <= mid) update2(o << 1, l, r, p);
  if (r > mid) update2(o << 1 | 1, l, r, p);
  t[o].sum = t[o << 1].sum + t[o << 1 | 1].sum;
  t[o].v = max(t[o << 1].v, t[o << 1 | 1].v);
}
long long query(int o, int l, int r) {
  if (l <= t[o].l && r >= t[o].r) {
    return t[o].sum;
  }
  long long ret = 0;
  int mid = (t[o].l + t[o].r) >> 1;
  if (l <= mid) ret += query(o << 1, l, r);
  if (r > mid) ret += query(o << 1 | 1, l, r);
  return ret;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  build(1, 1, n);
  while (m--) {
    int f;
    cin >> f;
    if (f == 1) {
      int x, y;
      cin >> x >> y;
      cout << query(1, x, y) << "\n";
    } else if (f == 2) {
      int x, y;
      long long z;
      cin >> x >> y >> z;
      update1(1, x, y, z);
    } else {
      int x;
      long long y;
      cin >> x >> y;
      update2(1, x, x, y);
    }
  }
  return 0;
}