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 SIZE = 100010;
struct SegementTree {
  int l, r;
  long long sum, add;
} tree[SIZE << 2];
int n, u, v, m;
void pushup(int p) {
  tree[p].sum = tree[p * 2].sum + tree[p * 2 + 1].sum;
  tree[p].add = max(tree[p * 2].add, tree[p * 2 + 1].add);
}
void build(int p, int l, int r) {
  tree[p].l = l, tree[p].r = r;
  if (l == r) {
    long long a;
    scanf("%lld", &a);
    tree[p].sum = tree[p].add = a;
    return;
  }
  int mid = (l + r) >> 1;
  build(p * 2, l, mid);
  build(p * 2 + 1, mid + 1, r);
  pushup(p);
}
void change1(int p, int x, int v) {
  if (tree[p].l == tree[p].r && tree[p].l == x) {
    tree[p].sum = v;
    tree[p].add = v;
    return;
  }
  int mid = (tree[p].l + tree[p].r) >> 1;
  if (x <= mid)
    change1(p * 2, x, v);
  else
    change1(p * 2 + 1, x, v);
  pushup(p);
}
void change2(int p, int l, int r, int t) {
  if (tree[p].add < t) return;
  if (tree[p].l == tree[p].r) {
    tree[p].sum %= t;
    tree[p].add %= t;
    return;
  }
  int mid = (tree[p].l + tree[p].r) >> 1;
  if (r <= mid) {
    change2(p * 2, l, r, t);
  } else if (l > mid) {
    change2(p * 2 + 1, l, r, t);
  } else {
    change2(p * 2, l, mid, t), change2(p * 2 + 1, mid + 1, r, t);
  }
  pushup(p);
}
long long ask(int p, int l, int r) {
  if (l == tree[p].l && r == tree[p].r) {
    return tree[p].sum;
  }
  int mid = (tree[p].l + tree[p].r) >> 1;
  if (r <= mid) {
    return ask(p * 2, l, r);
  } else if (l > mid) {
    return ask(p * 2 + 1, l, r);
  } else {
    return ask(p * 2, l, mid) + ask(p * 2 + 1, mid + 1, r);
  }
}
int main() {
  while (~scanf("%d %d", &n, &m)) {
    build(1, 1, n);
    while (m--) {
      int op, l, r, x;
      scanf("%d %d %d", &op, &l, &r);
      if (op == 1) {
        printf("%lld\n", ask(1, l, r));
      } else if (op == 2) {
        int s;
        scanf("%d", &s);
        change2(1, l, r, s);
      } else
        change1(1, l, r);
    }
  }
  return 0;
}