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>
int n, m, a[100005], t;
long long bit[100005];
std::pair<int, int> sgt2[400005];
void build2(int l, int r, int i) {
  int mid = (l + r) / 2;
  if (l == r) {
    sgt2[i] = std::make_pair(a[mid], mid);
    return;
  }
  build2(l, mid, i * 2);
  build2(mid + 1, r, i * 2 + 1);
  sgt2[i] = std::max(sgt2[i * 2], sgt2[i * 2 + 1]);
}
void update1(int x, int y) {
  y -= a[x];
  for (int i = x; i <= n; i += (i & (-i))) bit[i] += y;
}
void update2(int l, int r, int i, int x, int y) {
  int mid = (l + r) / 2;
  if (l == r) {
    sgt2[i] = std::make_pair(y, mid);
    return;
  }
  if (x <= mid)
    update2(l, mid, i * 2, x, y);
  else
    update2(mid + 1, r, i * 2 + 1, x, y);
  sgt2[i] = std::max(sgt2[i * 2], sgt2[i * 2 + 1]);
}
long long query(int x) {
  long long y = 0;
  for (int i = x; i; i -= (i & (-i))) y += bit[i];
  return y;
}
long long query1(int x, int y) { return query(y) - query(x - 1); }
std::pair<int, int> query2(int l, int r, int i, int x, int y) {
  if (x <= l && r <= y) return sgt2[i];
  std::pair<int, int> z = std::make_pair(0, 0);
  int mid = (l + r) / 2;
  if (l <= x && x <= mid || l <= y && y <= mid || x <= l && mid <= y)
    z = std::max(z, query2(l, mid, i * 2, x, y));
  if (mid + 1 <= x && x <= r || mid + 1 <= y && y <= r ||
      x <= mid + 1 && r <= y)
    z = std::max(z, query2(mid + 1, r, i * 2 + 1, x, y));
  return z;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) update1(i, a[i] + a[i]);
  build2(1, n, 1);
  for (int i = 0; i < m; i++) {
    scanf("%d", &t);
    if (t == 1) {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%I64d\n", query1(l, r));
    } else if (t == 2) {
      int l, r, x;
      scanf("%d %d %d", &l, &r, &x);
      std::pair<int, int> y = query2(1, n, 1, l, r);
      while (y.first >= x) {
        update1(y.second, y.first % x);
        update2(1, n, 1, y.second, y.first % x);
        a[y.second] = y.first % x;
        y = query2(1, n, 1, l, r);
      }
    } else {
      int k, x;
      scanf("%d %d", &k, &x);
      update1(k, x);
      update2(1, n, 1, k, x);
      a[k] = x;
    }
  }
}