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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct node {
  long long x, y;
};
node tre[4 * 200005];
void combine(long long n) {
  tre[n].x = tre[2 * n].x + tre[2 * n + 1].x;
  tre[n].y = max(tre[2 * n].y, tre[2 * n + 1].y);
}
void update(long long n, long long l, long long r, long long i, long long val) {
  if (l > i || r < i) {
    return;
  }
  if (l == i && i == r) {
    tre[n].x = tre[n].y = val;
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * n, l, mid, i, val);
  update(2 * n + 1, mid + 1, r, i, val);
  combine(n);
}
void mod(long long n, long long l, long long r, long long i, long long j,
         long long x) {
  if (tre[n].y < x) return;
  if (l > j || r < i) return;
  if (l == r) {
    tre[n].x = tre[n].y = tre[n].x % x;
    return;
  }
  long long mid = (l + r) / 2;
  mod(2 * n, l, mid, i, j, x);
  mod(2 * n + 1, mid + 1, r, i, j, x);
  combine(n);
}
long long query(long long n, long long l, long long r, long long i,
                long long j) {
  if (i > r || j < l) {
    return 0;
  }
  if (l >= i && r <= j) {
    return tre[n].x;
  }
  long long mid = (l + r) / 2;
  return query(2 * n, l, mid, i, j) + query(2 * n + 1, mid + 1, r, i, j);
}
void solve() {
  long long i, j, k, l, r, m, n, a, b, c, d, x, y, z, t;
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> x;
    update(1, 0, n - 1, i, x);
  }
  while (k--) {
    cin >> x;
    if (x == 1) {
      cin >> x >> y;
      cout << query(1, 0, n - 1, x - 1, y - 1) << endl;
    } else if (x == 2) {
      cin >> x >> y >> z;
      mod(1, 0, n - 1, x - 1, y - 1, z);
    } else {
      cin >> x >> y;
      update(1, 0, n - 1, x - 1, y);
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}