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, q, opt, l, r, x, val[100005];
struct segmentTree {
  int max[100005 << 2], left[100005 << 2], right[100005 << 2];
  long long sum[100005 << 2];
  int qmax(int a, int b) {
    if (val[a] > val[b]) return a;
    return b;
  }
  void update(int rt) {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    max[rt] = qmax(max[rt << 1], max[rt << 1 | 1]);
  }
  void build(int l, int r, int rt) {
    left[rt] = l;
    right[rt] = r;
    if (l == r) {
      scanf("%lld", &sum[rt]);
      max[rt] = l;
      val[l] = sum[rt];
      return;
    }
    build(l, (l + r) >> 1, rt << 1);
    build(((l + r) >> 1) + 1, r, rt << 1 | 1);
    update(rt);
  }
  int query_max(int l, int r, int rt) {
    if (left[rt] > r || right[rt] < l) return 0;
    if (l <= left[rt] && right[rt] <= r) return max[rt];
    return qmax(query_max(l, r, rt << 1), query_max(l, r, rt << 1 | 1));
  }
  void modify(int p, int v, int rt) {
    if (left[rt] == right[rt]) {
      sum[rt] = v;
      return;
    }
    if (p <= ((left[rt] + right[rt]) >> 1))
      modify(p, v, rt << 1);
    else
      modify(p, v, rt << 1 | 1);
    update(rt);
  }
  long long query_sum(int l, int r, int rt) {
    if (left[rt] > r || right[rt] < l) return 0;
    if (l <= left[rt] && right[rt] <= r) return sum[rt];
    return query_sum(l, r, rt << 1) + query_sum(l, r, rt << 1 | 1);
  }
} sgt;
int main() {
  scanf("%d%d", &n, &q);
  sgt.build(1, n, 1);
  while (q--) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d", &l, &r);
      printf("%lld\n", sgt.query_sum(l, r, 1));
    } else if (opt == 2) {
      scanf("%d%d%d", &l, &r, &x);
      int max;
      while (val[(max = sgt.query_max(l, r, 1))] >= x) {
        val[max] = val[max] % x;
        sgt.modify(max, val[max], 1);
      }
    } else if (opt == 3) {
      scanf("%d%d", &l, &x);
      val[l] = x;
      sgt.modify(l, x, 1);
    }
  }
  return 0;
}