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 maxn = 1 << 18;
int n, m;
int lazy[2 * maxn];
long long seg[2 * maxn];
int lim[maxn];
int qa, qb, qv, ix, x;
void update_mod(int rt, int a, int b) {
  if (a > qb || b < qa || lazy[rt] < x) return;
  if (a == b) {
    seg[rt] %= x;
    lazy[rt] = seg[rt];
    return;
  }
  int mid = (a + b) / 2;
  update_mod(2 * rt, a, mid);
  update_mod(2 * rt + 1, mid + 1, b);
  seg[rt] = seg[2 * rt] + seg[2 * rt + 1];
  lazy[rt] = max(lazy[2 * rt], lazy[2 * rt + 1]);
}
void update(int rt, int a, int b) {
  if (a > ix || b < ix) return;
  if (a == ix && b == ix) {
    lazy[rt] = qv;
    seg[rt] = qv;
    return;
  }
  int mid = (a + b) / 2;
  update(rt * 2, a, mid);
  update(rt * 2 + 1, mid + 1, b);
  seg[rt] = seg[rt * 2] + seg[rt * 2 + 1];
  lazy[rt] = max(lazy[rt * 2], lazy[rt * 2 + 1]);
}
long long query(int rt, int a, int b) {
  if (a > qb || b < qa) return 0;
  if (qa <= a && qb >= b) {
    return seg[rt];
  }
  int mid = (a + b) / 2;
  return (query(rt * 2, a, mid) + query(rt * 2 + 1, mid + 1, b));
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    ix = i + 1;
    scanf("%d", &qv);
    update(1, 1, n);
  }
  for (int i = 0; i < m; i++) {
    int qtype;
    scanf("%d", &qtype);
    if (qtype == 1) {
      scanf("%d%d", &qa, &qb);
      printf("%I64d\n", query(1, 1, n));
    } else if (qtype == 2) {
      scanf("%d%d%d", &qa, &qb, &x);
      update_mod(1, 1, n);
    } else if (qtype == 3) {
      scanf("%d%d", &ix, &qv);
      update(1, 1, n);
    }
  }
}