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;
int n, m;
struct Node {
  long long sum, mx;
} tree[100005 * 6];
void push_up(int node) {
  tree[node].mx = max(tree[node << 1].mx, tree[node << 1 | 1].mx);
  tree[node].sum = tree[node << 1].sum + tree[node << 1 | 1].sum;
}
void build(int left, int right, int node) {
  if (left == right) {
    scanf("%I64d", &tree[node].mx);
    tree[node].sum = tree[node].mx;
    return;
  }
  int mid = (left + right) >> 1;
  build(left, mid, node << 1);
  build(mid + 1, right, node << 1 | 1);
  push_up(node);
}
long long get_sum(int left, int right, int node, int L, int R) {
  if (L <= left && R >= right) {
    return tree[node].sum;
  }
  int mid = (left + right) >> 1;
  long long sum = 0;
  if (L <= mid) {
    sum += get_sum(left, mid, node << 1, L, R);
  }
  if (R > mid) {
    sum += get_sum(mid + 1, right, node << 1 | 1, L, R);
  }
  return sum;
}
void Update(int left, int right, int node, int L, int R, long long Mod) {
  if (tree[node].mx < Mod) {
    return;
  }
  if (left >= L && right <= R) {
    if (left == right) {
      tree[node].sum %= Mod;
      tree[node].mx %= Mod;
      return;
    }
    int mid = (left + right) >> 1;
    Update(left, mid, node << 1, L, R, Mod);
    Update(mid + 1, right, node << 1 | 1, L, R, Mod);
    push_up(node);
    return;
  }
  int mid = (left + right) >> 1;
  if (L <= mid) {
    Update(left, mid, node << 1, L, R, Mod);
  }
  if (R > mid) {
    Update(mid + 1, right, node << 1 | 1, L, R, Mod);
  }
  push_up(node);
}
void Update_one(int left, int right, int node, int p, long long val) {
  if (left == right) {
    tree[node].mx = val;
    tree[node].sum = val;
    return;
  }
  int mid = (left + right) >> 1;
  if (p <= mid) {
    Update_one(left, mid, node << 1, p, val);
  } else {
    Update_one(mid + 1, right, node << 1 | 1, p, val);
  }
  push_up(node);
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, n, 1);
  while (m--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      printf("%I64d\n", get_sum(1, n, 1, l, r));
    } else if (type == 2) {
      int l, r;
      long long Mod;
      scanf("%d%d%I64d", &l, &r, &Mod);
      Update(1, n, 1, l, r, Mod);
    } else {
      int p;
      long long val;
      scanf("%d%I64d", &p, &val);
      Update_one(1, n, 1, p, val);
    }
  }
  return 0;
}